Experiment 8 - Titration Curve

EXPERIMENT 8

TITRATION CURVE

GALVAN.KALAW

TITRATION

A common laboratory method of quantitative chemical analysis that is used to determine the unknown

concentration of a known reactant .

Also known as volumetric analysis (because volume measurements play a key role in titration).

A reagent, called the titrant or titrator, of known concentration (a standard solution), is added gradually to another solution of unknown concentration called analyte or titrand, until the chemical reaction between the two solutions is complete.

The term “equivalence point” means that the solutions have been mixed in exactly the right proportions according to the equation.

It is the point at which the acid has completely reacted with or been neutralized by the base.

The term “end point” is where the indicator changes color. It is not necessarily exactly the same as the equivalence point.

Indicators are substances that have distinctly different colors in acidic and basic media.

ACID – BASE TITRATIONS

Titration Curve would help determine whether a reaction can be used for titration or not. For acid – base reactions, a titration curve is obtained by plotting pH or pOH against volume of the titrant.

STRONG ACID – STRONG BASE

TITRATIONS

Volume of NaOH added (mL) pH

0 15 1.18

10 1.3715 1.620 1.9522 2.224 2.6925 726 11.2928 11.7530 11.9635 12.2240 12.3645 12.4650 12.52

0 10 20 30 40 50 600

2

4

6

8

10

12

14

pH

Volume of NaOH added (mL)

0.10 M NaOH solution to 25 mL of a 0.10 M HCl solution

Initial pH

HCl is a strong acid and completely dissociates, therefore,

[H3O+] = O.100 MpH = 1.0

After the addition of 10 mL of 0.10 M NaOH to 25 mL of 0.10 M HCl

10 mL X O.10 mol NaOH= 1mmol NaOH

25 mL X O.10 mol HCl= 2.5mmol HCl

H3O+ (aq) + OH- (aq) -> H2O (l)

2.5mmol 1mmol

-1mmol -1mmol

1.5mmol 0

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[H3O+] = 1.5 mmol / 35 ml = 0.043 M

pH = -log [H3O+] = -log 0.043

pH = 1.37


25 mL X O.10 mol NaOH= 2.5mmol NaOH


H3O+ (aq) + OH- (aq) -> H2O (l)

2.5mmol 2.5mmol

-2.5mmol -2.5mmol

0 0

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There is no more excess acid and the solution only contains water. At the equivalence point,

[H+] = [OH-]and the solution is neutral with a

pH = 7

H3O+ (aq) + OH- (aq) -> H2O (l)

2.5mmol 3.5mmol

-2.5mmol -2.5mmol

0 1mmol

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[OH-] = 1 mmol / 60 ml = 0.017 M

pOH = -log [OH-] = 1.78

pH = 14 – pOHpH = 12.22

WEAK ACID – STRONG BASE

TITRATIONS

Volume of NaOH added (mL) pH

0 2.875 4.14

10 4.5715 4.9220 5.3522 5.6124 6.1225 8.7226 10.2928 11.7530 11.9635 12.2240 12.3645 12.4650 12.52

0 10 20 30 40 50 600

2

4

6

8

10

12

14

pH


0.10 M NaOH solution to 25 mL of a 0.10 M HOAc solution

HOAc <-> H+ + OAc-

0.1 M 0 0

-x +x +x

0.1M – x +x +x

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Initial pH

Ka = x2 / 0.1 – x = 1.8x10-5

[H+] = 0.00134 MpH = 2.87

We can assume that x is negligible due to the very low value of Ka

After the addition of 10 mL of 0.10 M NaOH to 25 mL of 0.10 M HOAc

10 mL X O.10 mol NaOH= 1mmol NaOH

25 mL X O.10 mol HOAc= 2.5mmol HOAc

2.5mmol 1mmol

-1mmol -1mmol + 1mmol

1.5mmol 0 1mmol

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HOAc + OH- <-> OAc- + H2O

Using Henderson – Hasselbach Equation:

pH = pKa + log ([A-] / [HA])

pH = -log(1.8x10-5) +log (0.029/0.043)

pH = 4.57

2.5mmol 2.5mmol

-2.5mmol -2.5mmol + 2.5mmol

0 0 2.5mmol

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0.05 M

-x +x +x

0.05 – x +x +x

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OAc- + H2O <-> HOAc + OH-

Kb = Kw / Ka

Kb = 5.6 x 10 -10

Kb = x2 / 0.05 – x = 5.6x10-10

We can assume that x is negligible due to the very low value of Kb

[OH-] = 5.29 x 10-6 M

pOH = -log [OH-] = 5.28

pH = 14 – pOHpH = 8.72

2.5mmol 3.5mmol

-2.5mmol -2.5mmol +2.5mmol

0 1mmol 2.5mmol

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[OH-] = 1 mmol / 60 ml = 0.017 M

pOH = -log [OH-]

= 1.78

pH = 14 – pOHpH = 12.22

WEAK BASE – STRONG ACID TITRATIONS

Volume of HCl added (mL) pH

0 11.135 9.86

10 9.4415 9.0820 8.6622 8.3924 7.8825 5.2826 2.7028 2.2230 2.0035 1.7040 1.5245 1.4050 1.30

0 10 20 30 40 50 600

2

4

6

8

10

12

pH

Volume of HCl added (mL)

0.10 M HCl solution to 25 mL of a 0.10 M NH4OH solution

NH4OH <-> NH4+ + OH-

0.1 M 0 0

-x +x +x

0.1M – x +x +x

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Initial pH

Kb = x2 / 0.1 – x = 1.8x10-5

[OH-] = 0.00134 MpOH = 2.87

We can assume that x is negligible due to the very low value of Kb

pH = 14 – pOHpH = 11.13

After the addition of 10 mL of 0.10 M HCl to 25 mL of 0.10 M NH4OH

10 mL X O.10 mol HCl= 1mmol HCl

25 mL X O.10 mol NH4OH= 2.5mmol NH4OH

2.5mmol 1mmol

-1mmol -1mmol + 1mmol

1.5mmol 0 1mmol

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NH4OH + HCl <-> NH4Cl + H2O

Using Henderson – Hasselbach Equation:

pOH = pKb + log ([BH-] / [B])

pOH = -log(1.8x10-5) +log (0.029/0.043)

pOH = 4.57

pH = 14 – pOHpH = 9.43

2.5mmol 2.5mmol

-2.5mmol -2.5mmol + 2.5mmol

0 0 2.5mmol

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0.05 M

-x +x +x

0.05 – x +x +x

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NH4Cl + H2O <-> NH4OH + HCl

Ka = Kw / Kb

Ka = 5.6 x 10 -10

Ka = x2 / 0.05 – x = 5.6x10-10

We can assume that x is negligible due to the very low value of Ka

[H+] = 5.29 x 10-6 M

pH = -log [H+] pH = 5.28

2.5mmol 3.5mmol

-2.5mmol -2.5mmol +2.5mmol

0 1mmol 2.5mmol

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[H+] = 1 mmol / 60 ml = 0.017 M

pH = -log [H+]

pH = 1.78

TITRATION OF UNKNOWN

Volume of HCl added (mL) pH

0 3.452 4.34 4.686 4.987 5.098 5.219 5.35

10 5.5211 5.7612 6.12

12.5 6.7412.6 6.7912.7 7.0112.8 7.6

12.85 8.5312.95 9.9513.45 10.83

pH


0.10 M NaOH solution to 25 mL of an unknown weak acid solution

0 2 4 6 8 10 12 14 160

2

4

6

8

10

12

Concentration of unknown acid:

At equivalence point:

Mbase = 0.1 MVbase = 12.85mL

Macid = ?Vacid = 25mL

MbaseVbase = MacidVacid

0.1M x 13.85 = Macid x 25mL

Macid = 0.0514M

The pH is that of an acid and comparing the titration curve of the

unknown with the theoretical titration curves, it can be concluded that or unknown is a weak acid. It

has a less steep pH due to the presence of a buffer.

DISCUSSION

Part A.1.

In weak acid & strong base titration , the pH change is not so steep due to the presence of a buffer. But, the pH change is seen when it reaches the equivalence point, wherein the concentrations of the acid & base are the same. As we add more base, the pH increases.

To get the volume or concentration at equivalence point. We use the equation:

MbaseVbase = MacidVacid

Part A.2.

In strong acid & strong base titration, the initial pH of the acid is lower than that of A.1. since it is stronger. The pH change is steeper. Theoretically, the equivalence point is always 7. As we add more base, the pH increases as well.

Part A.3.

In weak base & strong acid titration, the initial pH is greater than 7 since it is a base. As we add more acid, the pH decreases. As we reach the equivalence point, the base undergoes hydrolysis which makes the equivalence point less than 7. Afterwards, the acid is the one tested for pH.

CONCLUSION&

RECOMMENDATION

There are four important points in titration:1. Start of titration2. Before equivalence point3. At equivalence point4. After equivalence point

Before the titration reaches the equivalence point, the pH of the solution depends on the amount of the substance not yet neutralized by the titrant. After reaching the equivalence point, pH levels depend on the excess amount of the titrant in the solution.

Determining the pH at equivalence point depends of the substances used in titration. For a strong acid-strong base titration, there is no need to compute for the pH level because it is always seven. There are no cations or anions that undergo hydrolysis which has an effect on its pH level.

Otherwise, take hydrolysis in consideration.

If titrating without a magnetic stirrer, DO NOT FORGET TO SWIRL!This may affect the pH of the solution.

Observe proper handling of reagents especially acids and instruments in titrating especially the burette to avoid accidents.Acids are harmful and burettes are fragile and EXPENSIVE! Make sure you handle it with care especially if PYREX.

Handle the pH meter with UTMOST care.It’s NEW!

Sources:

Chem 18 ModuleGeneral Chemistry by Chang

Experiment 8 - Titration Curve

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Transcript of Experiment 8 - Titration Curve