Acid-Base Titrations
description
Transcript of Acid-Base Titrations
![Page 1: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/1.jpg)
Acid-Base Titrations
Section 17.3
![Page 2: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/2.jpg)
Introduction
• Definition:– In an acid-base titration, a solution containing a
known concentration of a base is slowly added to an acid.
– An indicator is used to signal the equivalence point of the titration.• This is the point at which stoichiometrically equivalent
amounts of acid and base have been mixed.– A pH meter can also be used to find the
equivalence point.
![Page 3: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/3.jpg)
Introduction
• The typical titration apparatus includes:– a buret to hold the titrant– a beaker to hold the analyte– a pH meter to measure the pH
![Page 4: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/4.jpg)
Introduction
• In this section, we will be looking at a series of titrations in detail to understand why acids and behave the way they do.– Strong acid-strong base titration– Weak acid-strong base titration– Polyprotic acid-strong base titration
![Page 5: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/5.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
![Page 6: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/6.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:1. Initial pH
2. Initial pH to eq. point
3. Equivalence point
4. After eq. point
![Page 7: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/7.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:1. Initial pH
The pH of the solution is determined by the concentration of the strong acid.
![Page 8: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/8.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
As base is added, pH increases slowly and then rapidly. The pH is determined by the concentration of the acid that is not neutralized.
2. Initial pH to eq. point
![Page 9: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/9.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
At the equivalence point, [OH−] = [H+]. The pH = 7.00.
3. Equivalence point
![Page 10: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/10.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
We divide the curve into four regions:
As more base is added, pH increases rapidly and then slowly. pH is determined by the concentration of the excess base.
4. After eq. point
![Page 11: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/11.jpg)
Strong Acid-Strong Base Titrations
• The titration curve of a strong acid-strong base titration has the following shape.
Let’s see how this works in practice.
![Page 12: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/12.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL
![Page 13: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/13.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL
![Page 14: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/14.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point.
![Page 15: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/15.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLThis is between the initial point and the equivalence point. pH is determined by the amount of acid that has not been neutralized.
![Page 16: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/16.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLTherefore, we need to determine the number of mols of acid remaining, nacid, and the total volume, Vtotal, of the solution. (Remember, adding the NaOH increases the total volume.
![Page 17: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/17.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i
![Page 18: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/18.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = (0.100 M)(0.0500 L)
![Page 19: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/19.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 mol
![Page 20: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/20.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = Mbase,added × Vbase,added
![Page 21: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/21.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = (0.100 M)(0.0490 L)
![Page 22: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/22.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 mol
![Page 23: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/23.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = nacid,i − nbase,added
![Page 24: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/24.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = (5.00 − 4.90) × 10−3 mol
![Page 25: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/25.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = 0.10 × 10−3 mol
![Page 26: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/26.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,i = Macid,i × Vacid,i = 5.00 × 10−3 molnbase,added = 4.90 × 10−3 molnacid,remaining = 1.0 × 10−4 mol
![Page 27: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/27.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 mol
![Page 28: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/28.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase
![Page 29: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/29.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0500 L + 0.0490 L
![Page 30: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/30.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0990 L
![Page 31: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/31.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L
![Page 32: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/32.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = nacid,remaining/Vtotal
![Page 33: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/33.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = (1.0 × 10−4 mol)/(0.0990 L)
![Page 34: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/34.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 M
![Page 35: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/35.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = −log[H+]
![Page 36: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/36.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = −log(1.0 × 10−3)
![Page 37: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/37.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLnacid,remaining = 1.0 × 10−4 molVtotal = 0.0990 L[H+] = 1.0 × 10−3 MpH = 3.00
![Page 38: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/38.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLpH = 3.00
![Page 39: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/39.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mL
![Page 40: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/40.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.a) 49.0 mLb) 51.0 mL
![Page 41: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/41.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.
b) 51.0 mL
![Page 42: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/42.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mL
![Page 43: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/43.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point.
![Page 44: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/44.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up.
![Page 45: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/45.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLThis is beyond the equivalence point. All of the strong acid has been used up. The pH is determined by the excess base that has been added.
![Page 46: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/46.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLWe already determined nacid,i = 5.00 × 10−3 mol
![Page 47: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/47.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 mol
![Page 48: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/48.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = Mbase,added × Vbase,added
![Page 49: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/49.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = (0.100 M)(0.0510 L)
![Page 50: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/50.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 mol
![Page 51: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/51.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = nbase,added − nacid,i
![Page 52: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/52.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = (5.10 − 5.00) × 10−3 mol
![Page 53: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/53.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnacid,i = 5.00 × 10−3 molnbase,added = 5.10 × 10−3 molnbase,remaining = 0.10 × 10−3 mol
![Page 54: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/54.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 mol
![Page 55: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/55.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase
![Page 56: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/56.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.0500 L + 0.0510 L
![Page 57: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/57.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = Vacid + Vbase = 0.1010 L
![Page 58: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/58.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L
![Page 59: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/59.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = nbase,remaining/Vtotal
![Page 60: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/60.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = (1.0 × 10−4 mol)/(0.1010 L)
![Page 61: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/61.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 M
![Page 62: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/62.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = −log[OH−]
![Page 63: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/63.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = −log(9.9 × 10−4)
![Page 64: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/64.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00
![Page 65: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/65.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00⇒
![Page 66: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/66.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpOH = 3.00 pH = 14.00 − 3.00 = 11.00⇒
![Page 67: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/67.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLnbase,remaining = 1.0 × 10−4 molVtotal = 0.1010 L[OH−] = 9.9 × 10−4 MpH = 11.00
![Page 68: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/68.jpg)
Sample Exercise 17.6 (page 731)
Calculate the pH when the following quantities of 0.100 M NaOH solution have been added to 50.0 mL of 0.100 M HCl solution.b) 51.0 mLpH = 11.00
![Page 69: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/69.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
![Page 70: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/70.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
1. The initial pH is determined by the Ka of the acid.
![Page 71: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/71.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
2. To determine the pH from the initial point to the eq. point, we first neutralize the weak acid and then use the Henderson-Hasselbach equation.
![Page 72: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/72.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
3. At the eq. point, we have no HX, only X−. We need to use the Kb value to find pH.
![Page 73: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/73.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
4. Beyond the eq. point, we use the excess base to calculate pH.
![Page 74: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/74.jpg)
Weak Acid-Strong Base Titrations
• The curve for a weak acid-strong base titration is similar to that for a strong acid-weak base.
Let’s see how this works in practice.
![Page 75: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/75.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
![Page 76: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/76.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
![Page 77: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/77.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2O
![Page 78: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/78.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid
![Page 79: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/79.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid = (0.100 M)(0.0500 L)
![Page 80: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/80.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = Macid × Vacid = 5.00 × 10−3 mol
![Page 81: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/81.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 mol
![Page 82: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/82.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase
![Page 83: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/83.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase = (0.100 M)(0.0450 L)
![Page 84: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/84.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = Mbase × Vbase = 4.50 × 10−3 mol
![Page 85: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/85.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
First, calculate the concentrations of materials before neutralization reaction.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = 4.50 × 10−3 mol
![Page 86: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/86.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial,acid = 5.00 × 10−3 molnbase = 4.50 × 10−3 mol
![Page 87: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/87.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
![Page 88: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/88.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol
![Page 89: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/89.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Oninitial 5.0 × 10−3 mol 4.5 × 10−3 mol 0.0 mol
nchange −4.5 × 10−3 mol −4.5 × 10−3 mol +4.5 × 10−3 mol
nfinal 5.0 × 10−4 mol 0.0 mol 4.5 × 10−3 mol
![Page 90: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/90.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol
![Page 91: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/91.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = nacid,final/Vtotal
![Page 92: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/92.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = (5.0 × 10−4 mol)/(0.0950 L)
![Page 93: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/93.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2Onacid,final = 5.0 × 10−4 mol[acid] = 5.3 × 10−3 M
![Page 94: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/94.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = nbase,final/Vtotal
![Page 95: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/95.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = (4.50 × 10−3 mol)/(0.0950 L)
![Page 96: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/96.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M[base] = 4.74 × 10−2 M
![Page 97: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/97.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 M
![Page 98: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/98.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −logKa
![Page 99: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/99.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = −log(1.8 × 10−5)
![Page 100: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/100.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2O[acid] = 5.3 × 10−3 M [base] = 4.74 × 10−2 MpKa = 4.74
![Page 101: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/101.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2OpH = pKa + log([base]/[acid)
![Page 102: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/102.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + log([4.74 × 10−2 M]/[5.3 × 10−3 M])
![Page 103: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/103.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + log(9.00)
![Page 104: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/104.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2OpH = 4.74 + 0.954
![Page 105: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/105.jpg)
Sample Exercise 17.7 (page 735)
• Calculate the pH of the solution formed when 45.0 mL of 0.100 M NaOH is added to 50.0 mL of 0.100 M CH3COOH.
Next, use the values to determine the concentrations after the neutralization.
CH3COOH + OH− CH➙ 3COO− + H2OpH = 5.69
![Page 106: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/106.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
![Page 107: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/107.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
At the equivalence point, all of the weak acid is converted to its conjugate weak base.
![Page 108: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/108.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
At the equivalence point, all of the weak acid is converted to its conjugate weak base. This means that we will be using the base hydrolysis expression to find [OH−], pOH, and pH.
![Page 109: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/109.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
The number of mols of acetate in solution at the equivalence point is equal to the number of mols of acetic acid at the start of the titration.
![Page 110: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/110.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid
![Page 111: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/111.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = (0.100 M)(0.0500 L)
![Page 112: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/112.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = ninitial,acid = 5.00 × 10−3 mol
![Page 113: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/113.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 mol
![Page 114: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/114.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:
![Page 115: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/115.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = (nbase,eq. pt.)/(Vtotal)
![Page 116: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/116.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = (5.00 × 10−3 mol)/(0.100 L)
![Page 117: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/117.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
nbase,eq. pt. = 5.00 × 10−3 molThe concentration of the acetate is given by:[base] = 5.00 × 10−2 M
![Page 118: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/118.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M
![Page 119: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/119.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MNext, we do an i-c-e table on the base hydrolysis.
![Page 120: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/120.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
![Page 121: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/121.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0
![Page 122: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/122.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +x
![Page 123: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/123.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 MCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x
![Page 124: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/124.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xCH3COO− + H2O CH➙ 3COOH + OH−
i 5.00 × 10−2 0.0 0.0c −x +x +xe 5.00 × 10−2 − x x x
![Page 125: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/125.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = (1.0 × 10−14)/(1.8 × 10−5)
![Page 126: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/126.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = Kw/Ka = 5.6 × 10 −10
![Page 127: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/127.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10
![Page 128: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/128.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = ([acid][OH−])/[base]
![Page 129: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/129.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)
![Page 130: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/130.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½
![Page 131: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/131.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6
![Page 132: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/132.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[base] = 5.00 × 10−2 M, [acid] = [OH−] = xKb = 5.6 × 10 −10 = (x2)/(5.00 × 10−2)x2 = (5.6 × 10 −10)(5.00 × 10−2) = 2.8 × 10−11
x = (2.8 × 10−11)½ = 5.3 × 10−6 = [OH−]
![Page 133: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/133.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 M
![Page 134: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/134.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−]
![Page 135: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/135.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6)
![Page 136: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/136.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = −log[OH−] = −log(5.3 × 10−6) = 5.28
![Page 137: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/137.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28
![Page 138: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/138.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH
![Page 139: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/139.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28
![Page 140: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/140.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
[OH−] = 5.3 × 10−6 MpOH = 5.28pH = 14.00 – pOH = 14.00 − 5.28 = 8.72
![Page 141: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/141.jpg)
Sample Exercise 17.8 (page 735)
• Calculate the pH at the equivalence point in the titration of 50.0 mL of 0.100 M CH3COOH with 0.100 M NaOH.
pH = 8.72
![Page 142: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/142.jpg)
• The pH of the titration at the equivalence point depends on the type of acids and bases being titrated.– Strong acid-Strong base: eq. pt. = 7.0– Weak acid-Strong base: eq. pt. > 7.0– Strong acid-Weak base: eq. pt. < 7.0
![Page 143: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/143.jpg)
Titrations of Polyprotic Acids
• When weak acid contain more than one ionizable H atom, as in phosporous acid, H3PO3, reaction with OH− occurs in a series of steps.– H3PO3(aq) + H2O(l) H➙ 2PO3
−(aq) + H3O+(aq)– H2PO3
−(aq) + H2O(l) HPO➙ 32−(aq) + H3O+(aq)
– HPO32−(aq) + H2O(l) PO➙ 3
3−(aq) + H3O+(aq)
![Page 144: Acid-Base Titrations](https://reader033.fdocuments.net/reader033/viewer/2022061509/5681654d550346895dd7c782/html5/thumbnails/144.jpg)
Titrations of Polyprotic Acids
• When the neutralization steps are sufficiently separated, the substance exhibits multiple equivalence points.