CHEMISTRY - GILBERT 5E CH.16 - ADDITIONAL ...lightcat-files.s3.amazonaws.com › packets ›...

! www.clutchprep.com

!

CHEMISTRY - GILBERT 5E

CH.16 - ADDITIONAL AQUEOUS EQUILIBRIA

CONCEPT: CLASSIFICATION AND IDENTIFICATION OF BUFFERS

Solutions which contain a _________________ acid and its __________________ base are called buffer solutions because they resist drastic changes in pH.

• They resist drastic changes in pH by keeping ___________ and ___________ constant.

• Adding a small amount of STRONG BASE and, the pH ____________, but not by much because the

________________ neutralizes the STRONG BASE added.

• Adding a small amount of STRONG ACID and, the pH ____________, but not by much because the

________________ neutralizes the STRONG ACID added.

PRACTICE 1: Which one of the following combinations does not create a buffer?

a) HC2H3O2 and K C2H3O2

b) H2SO3 and NaHSO3

c) H3PO4 and NaH2PO4

d) HNO3 and KNO3

e) NH4Cl and NH3

PRACTICE 2: Which of the following combinations can result in the formation of a buffer?

a) HF and HI

b) HC2H3O2 and NH3

c) CH3CH2NH2 and CH3CH2NH3+

d) NaCl and NaOH



CONCEPT: CREATING A BUFFER

There are 3 ways to form a buffer:

1) Mixing a ______________ acid and its ______________ base.

2) Mixing a ______________ acid and a ______________ base.

3) Mixing a ______________ acid and a ______________ base.



PRACTICE: CREATING A BUFFER

EXAMPLE: Which of the following combinations can result in the formation of a buffer?

a) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of NaOH.

b) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of HCl.

c) 0.01 moles HClO (hypochlorous acid) and 0.05 moles of NH3.

d) 0.01 moles HClO (hypochlorous acid) and 0.001 moles of NaOH

PRACTICE 1: Which of the following combinations can result in the formation of a buffer?

a) 50 mL of 0.10 M HF with 50 mL of 0.10 M NaOH.

b) 50 mL of 0.10 M HNO2 with 25 mL of 0.10 M Ca(OH)2.

c) 50 mL of 0.10 M CH3CO2H with 60 mL of 0.10 M NaOH.

d) 50 mL of 0.10 M HF with 30 mL of 0.10 M NaOH.

PRACTICE 2: A buffer solution is comprised of 50.0 mL of a 0.100 M HC2H3O2 and 60.0 mL of a 0.100 M NaC2H3O2. Which

of the following actions would completely destroy the buffer?

a) Adding 0.003 mol HC2H3O2

b) Adding 0.007 mol Ca(C2H3O2)2

c) Adding 0.005 mol NaOH

d) Adding 0.004 mol HCl

e) Adding 0.001 mol HCl



CONCEPT: CALCULATING THE pH OF BUFFERS

We learned that whenever we had a(n) ____________ acid or base we were supposed to use our favorite friend the

_____________ Chart in order to calculate the pH or pOH.

Now, whenever we have a buffer solution we can skip it and use the ___________________________________ Equation.

Buffer Equation:

pH = pKa + log (conjugate base)(weak acid)

EXAMPLE 1: What is the pH of a solution consisting of 2.75 M sodium phenolate (C6H5ONa) and 3.0 M phenol (C6H5OH).

The Ka of phenol is 1.0 x 10-10.

PRACTICE: Calculate the pH of a solution formed by mixing 200 mL of a 0.400 M C2H5NH2 solution with 350 mL of a 0.450

M C2H5NH3+ solution. (Kb of C2H5NH2 is 5.6 x 10-4).



PRACTICE: CALCULATING THE pH OF BUFFERS (PART 1)

EXAMPLE 1: What is the buffer component concentration ratio, Pr–

HPr, of a buffer that has a pH of 5.11. (The Ka of HPr is

1.30 x 10-5).

EXAMPLE 2: Over what pH range will an oxalic acid (H2C2O4) / sodium oxalate (NaHC2O4) solution work most effectively?

The acid dissociation constant of oxalic acid is 6.0 x 10-2.

a) 0.22 – 2.22 b) 1.00 – 3.00 c) 0.22 – 1.22 d) 2.0 – 4.0

PRACTICE: Determine how many grams of sodium acetate, NaCH3CO2 (MW: 82.05 g/mol), you would mix into enough

0.065 M acetic acid CH3CO2H (MW: 60.05 g/mol) to prepare 3.2 L of a buffer with a pH of 4.58. The Ka is 1.8 x 10-5.



PRACTICE: CALCULATING THE pH OF BUFFERS (PART 2)

EXAMPLE: Which weak acid-conjugate base combination would be ideal to form a buffer with a pH of 4.74.

a) Cyanic acid and Potassium cynate (Ka = 4.9 x 10-10)

b) Benzoic acid and Lithium benzoate (Ka = 6.3 x 10-5)

c) Acetic acid and Sodium acetate (Ka = 1.7 x 10-5)

d) Ammonium chloride and Ammonia (Ka = 5.56 x 10-10)

e) Formic acid and Cesium formate (Ka = 1.7 x 10-4)

PRACTICE: A buffer solution is made by combining a weak acid with its conjugate salt. What will happen to the pH if the solution is diluted to one-fourth of its original concentration?

a) The pH will increase.

b) The pH will decrease.

c) The pH will remain constant.

d) The solution will become more neutral.



CONCEPT: pH TITRATION CURVES

The shape of a pH titration curve makes it possible to identify the equivalence point, the point at which _________________

of acid and base are mixed together.

At the equivalence point, the pH of a strong acid and strong base is ___________ 7.

At the equivalence point, the pH of a strong acid and weak base is ___________ 7.

At the equivalence point, the pH of a weak acid and strong base is ___________ 7.



PRACTICE: pH TITRATION CURVES (CALCULATIONS 1)

EXAMPLE: The following questions refer to the titration curve given below. a) The titration curve shows the titration of a strong acid a weak acid a strong base a weak base with a strong base with a strong base with a strong acid with a strong acid b) Which point on the titration curve represents a region where a buffer solution has formed? point A point B point C point D c) Which point on the titration curve represents the equivalence point? point A point B point C point D d) Which of the following would be the best indicator to use in the titration? erythrosin B methyl red bromthymol blue o-cresonphthalein pKa = 2.9 pKa = 5.4 pKa = 6.8 pKa = 9.0




EXAMPLE 1: The acid form of an indicator is red and its anion is blue. The Ka value for this indicator is 10-9. What will be

the approximate pH range over which this indicator changes color?

a) 3-5 b) 4-6 c) 5-7 d) 8-10 e) 9-11

PRACTICE : What will be the color of the indicator in the above question in a solution that has a pH of 6?

EXAMPLE 2: Consider the titration of 100.0 mL of 0.016 M H2SO4 with 0.400 M NaOH at the equivalence point. How many

many milliters of 0.400 M NaOH are required to reach the equivalence point?




EXAMPLE: Consider the titration of 40.0 mL of 0.0800 M HCl with 0.0160 M Ca(NH2)2.

a) How many milliliters of 0.0160 M Ca(NH2)2 are required to reach the equivalence point?

b) What is the pH of this solution?

PRACTICE: Consider the titration of 60.0 mL of 0.200 M a H3PO3 solution with 0.350 M potassium hydroxide, KOH solution.

How many milliliters of base would be required to reach each of its equivalence points?



CONCEPT: WEAK ACID & STRONG BASE TITRATIONS

In the past, we reacted WEAK Acids or WEAK Bases with ___________ and used a(n) _________________ Chart.

• Remember in this case the units in this chart are in _______________.

Now we will react WEAK Acids with ______________ bases.

• When you react a WEAK species with a ____________ species you have to use a(n) _________________ Chart.

• In this case the units must be in _______________.

HNO2 NaOHWeak Acid Strong Base

Before Equivalence Point

HNO2 NaOHWeak Acid Strong Base Conjugate Base

NaNO2 H2O

Initial

Change

Final

0.250 moles

•_______________ and _______________ will be present at the end. •Use the Henderson Hasselbalch Equation.

pH = pKa + log Conjugate BaseWeak Acid

⎛

⎝⎜

⎞

⎠⎟

After Equivalence Point

Initial

Change

Final

0.250 moles

At Equivalence Point

Initial

Change

Final

0.250 moles

• _______________ will be present at the end.

[SB]= moles leftTotal Liters

∴ pOH = − log[SB] ∴ pH =14− pOH

•Only _______________ will be present at the end. •Use an ICE Chart to find pH.

[CB]= moles leftTotal Liters

∴ Kb =x2

[CB]


NaNO2 H2O


NaNO2 H2O



PRACTICE: WEAK ACID & STRONG BASE TITRATIONS CALCULATIONS 1

EXAMPLE: Consider the titration of 75.0 mL of 0.0300 M H3C3O3 (Ka = 4.1 X 10-3) with 12.0 mL of 0.0450 M KOH.

Calculate the pH.

PRACTICE: In order to create a buffer, 7.510 g of sodium cyanide is mixed with 100.0 mL of 0.250 M hydrocyanic acid, HCN. What is the pH of the buffer solution after the addition of 175.0 mL of 0.300 M NaH?



PRACTICE: WEAK ACID & STRONG BASE TITRATIONS CALCULATIONS 2

EXAMPLE: Consider the titration of 75.0 mL of 0.60 M HNO2 with 0.100 M NaOH at the equivalence point. What would be

the pH of the solution at the equivalence point? The Ka of HNO2 is 4.6 x 10-4.



CONCEPT: WEAK BASE & STRONG ACID TITRATIONS

Now we will react WEAK Bases with STRONG Acids.

• Remember when you react a WEAK species with a ___________ species you have to use a(n) __________ Chart.

• Again, in this case the units must be in _______________.

NH3 HBr

Before Equivalence Point

Initial

Change

Final

0.250 moles

•_______________ and _______________ will be present at the end. •Use the Henderson Hasselbalch Equation.

pH = pKa + log Conjugate BaseWeak Acid

⎛

⎝⎜

⎞

⎠⎟

After Equivalence Point

Initial

Change

Final

0.250 moles

At Equivalence Point

Initial

Change

Final

0.250 moles

• _______________ will be present at the end.

•Only _______________ will be present at the end. •Use an ICE Chart to find pH.

NH3 HBrConjugate Base Strong Acid Weak Acid

NH4+ Br –


NH4+ Br –


NH4+ Br –

[SA]= moles leftTotal Liters

∴ pH = − log[SA]

[WA]= moles leftTotal Liters

∴ Ka =x2

[CB]



PRACTICE: WEAK BASE & STRONG ACID TITRATIONS CALCULATIONS

EXAMPLE: A buffer contains 167.2 mL of 0.25 M propanoic acid, CH3CH2COOH, with 138.7 mL of 0.42 M sodium propanoate, CH3CH2COONa. Find the pH after the addition of 150.2 mL of 0.56 M HCl. The Ka of CH3CH2COOH is 1.3 x

10-5.

PRACTICE: Calculate the pH of the solution that results from the mixing of 75.0 mL of 0.100 M NaC2H3O2 and 75.0 mL of

0.150 M HC2H3O2 with 0.0025 moles of HBr. Ka of HC2H3O2 is 1.8 x 10-5.



CONCEPT: STRONG ACID & BASE TITRATIONS

Whenever we had a STRONG ACID or STRONG BASE we ____________ use an ICE CHART.

Now, whenever you titrate a STRONG ACID with STRONG BASE you ____________ use an ICF CHART.

EXAMPLE: Calculate the pH of the solution resulting from the titration of 75.0 mL of 0.100 M HBrO4 with 55.0 mL of 0.100

M NaNH2.

PRACTICE: Calculate the pH of the solution resulting from the mixing of 175.0 mL of 0.250 M HNO3 with 75.0 mL of 0.200

M Ba(OH)2.



CONCEPT: DIPROTIC & POLYPROTIC BUFFERS

A diprotic or polyprotic buffer can be approached in a way similar to monoprotic buffers. The key difference is that multiple pKa values will be involved.

For Monoprotic Buffers

pH = pKa + log (conjugate base)(weak acid)

For Diprotic Buffers

H2A HA– A2–

pH =

pH =

For Polyprotic Buffers

H2A– HA2 – A3 –H3A

pH =

pH =

pH =

HClO + NaClO (aq)Weak oxyacid Conjugate base

(Has 1 less Hydrogen)



PRACTICE: DIPROTIC & POLYPROTIC BUFFERS CALCULATIONS 1

EXAMPLE 1: Calculate the pH of 100 mL of a 0.25 M H2CO3 when 70.0 mL of 0.25 M NaOH are added. Ka1 = 4.3 x 10-7 and Ka2 = 5.6 x 10-11.

EXAMPLE 2: Calculate the pH of 75.0 mL of a 0.10 M of phosphorous acid, H3PO3, when 80.0 mL of 0.15 M NaOH are added. Ka1 = 5.0 x 10-2, Ka2 = 2.0 x 10-7.



CONCEPT: CALCULATING SOLUBILITY FROM Ksp

__________________________ was the maximum amount of solute that could successfully dissolved in a solvent.

In this section we learn that associated with any solid is a ___________ value, which stands for the solubility

product constant.

The larger the solubility product constant then the _________ soluble the solid is in a solvent.

The smaller the solubility product constant then the _________ soluble the solid is in a solvent.

EXAMPLE: Consider the following compounds. Which has the highest concentration of OH- ions?

a) Co(OH)2 Ksp = 1.3 x 10-15

b) Cu(OH)2 Ksp = 2.2 x 10-20

c) Ni(OH)2 Ksp = 6.0 x 10-16

d) Fe(OH)2 Ksp = 4.1 x 10-15

e) Zn(OH)2 Ksp = 3.0 x 10-16

EXAMPLE: The solubility of silver sulfate, Ag2SO4, is 0.025 M at 25oC. Calculate its solubility product constant, Ksp.



PRACTICE: CALCULATING SOLUBILITY FROM Ksp (CALCULATIONS 1)

EXAMPLE: Find the solubility of CoCl3 (Ksp = 2.8 x 10-13) in

a) pure water

b) 0.20 M NaCl.




EXAMPLE 1: What is the molar solubility of Fe(OH)3 (s) in a solution that is buffered at pH 3.50 at 25°C? The Ksp of

Fe(OH)3 is 6.3 x 10–38 at 25°C.

EXAMPLE 2: Find the pH of a saturated solution of Aluminum hydroxide, Al(OH)3. The Ksp of Al(OH)3 is 1.9 x 10-10.




EXAMPLE 1: Will a precipitate form when 0.150 L of 0.100 M Pb(C2H3O2)2 and 0.100 L of 0.20 M KCl are mixed? The Ksp value of PbCl2 is 1.2 x 10-5.

EXAMPLE 2: Two flasks containing 7.12 x 10-6 M silver nitrate, AgNO3, and 8.33 x 10-4 M strontium bromide, SrBr2

respectively are mixed together into a larger flask. If the Ksp of silver bromide, AgBr, is 5.35 x 10-13 what will happen?

a) A precipitate of Sr(NO3)2 will form. b) The solution is saturated but will remain clear. c) The solution is not saturated and no precipitate will form. d) A precipitate of silver bromide will form, with silver and bromide ions as well. e) A precipitate of silver bromide will form.




EXAMPLE 1: A saturated solution of AlCl3 contains 5.15 x 10-4 M chloride ions. What is the solubility product constant, Ksp?

EXAMPLE 2: Ca3(PO4)2 has a Ksp of 5.8 x 10-32. What is the molar solubility of calcium ions in a saturated solution of

Ca3(PO4)2?




EXAMPLE 1: The solubility of CaF2 is 53.2 g/mL at 25 oC. Calculate its solubility product constant, Ksp.

EXAMPLE 2: The concentration of chloride and iodide ions present in the solution are [Cl –] = 0.073 M and [I –] = 0.060 M.

The use of silver, Ag, can be used to precipitate these two ions. If the Ksp value of AgCl is 1.77 x 10-10 and AgI is

8.51 x 10-17.

a) Which ion will precipitate out first? b) At what concentration will this ion from Part A begin to precipitate out?



CHEMISTRY - GILBERT 5E CH.16 - ADDITIONAL ...lightcat-files.s3.amazonaws.com › packets ›...

Documents

Transcript of CHEMISTRY - GILBERT 5E CH.16 - ADDITIONAL ...lightcat-files.s3.amazonaws.com › packets ›...