Generate a Curve for the Titration of 50.00 Ml of 0.1000 M Acetic Acid with 0.1000 M Sodium

Example 3

Generate a curve for the titration of 50.00 mL of 0.1000 M acetic acid with 0.1000 M sodium hydroxide.

Initial pH

Calculate the pH of a 0.1000 M solution of HOAc

Ka = [H3O+][OAc-]/[HOAc] = x2/[HOAc]

[H3O+] = x = (Ka[HOAc])1/2 = (1.75 X 10-5 X 0.100)1/2 = 1.32 X 10-3 M

pH = -log (1.32 X 10-3) = 2.88

pH after Addition of 5.00 mL of Reagent

A buffer solution consisting of NaOAc and HOAc has been produced.

[HAOc] = (50.00 mL X 0.100 M - 5.00 mL X 0.100 M)/55.00 = 4.500/55 M

[OAc-] = (5.00 mL X 0.100 M)/55 = 0.500/55.00 M

Ka = [H3O+] (0.500/55.00)/(4.500/55.00) = 1.75 X 10-5

[H3O+] = 1.58 X 10-4 M

pH = 3.80

pH after Addition of 25.00 mL of Reagent;(half-Neutralization)

[HAOc] = (50.00 mL X 0.100 M - 25.00 mL X 0.100 M)/75.00 = 2.500/75.00 M

[OAc-] = (25.00 mL X 0.100 M)/75.00 = 2.500/75.00 M

Ka = [H3O+] (2.500/75.00)/(2.500/75.00) = [H3O+] =1.75 X 10-5

pH = pKa = 4.76

Equivalence Point pH

At the equivalence point, all the acetic acid has been converted to sodium acetate. The solution is therefore similar to one formed by dissolving that salt in water. In the present example, the NaOAc concentration is 0.0500 M.

OAc- + H2O ↔ HOAc + OH-

[OAc-] = 0.0500 - [OH-] ≈ 0.0500 M

Kb= Kw/Ka = 1X10-14/1.75 X 10-5 = 5.71 X 10-10

Kb = [HAOc][OH-]/[OAc-] = x2/[OAc-] = 5.71 X 10-10

[OH-] = x = (0.0500 X 5.71 X 10-10)1/2 = 5.34 X 10-6 M

pH = 14.00 - (-log 5.34 X 10-6 ) = 8.73

pH after Addition of 50.01 mL of Base

After equivalence, both excess strong base and the acetate ion (weak base) are present. The contribution from the acetate ion is small. However, because the excess of strong base represses the reaction of acetate with water, the contribution from the acetate is even smaller.

[OH-] ≈ [NaOH] = (50.01 mL X 0.1000 M - 50.00 mL X 0.1000 M)/100.01mL

= 1.00 X 10-5 M

pH = 14.00 - [-log (1.00 X 10-5)] = 9.00

Additional data for this titration (second column) is compared with a more dilute titration in the table below:

These data are drawn in the next figure:

Curve for the titration of acetic acid with sodium hydroxide. Curve A: 0.1000 M acid with 0.1000 M base. Curve B: 0.001000 M acid with 0.001000 M base.

The Effect of Concentration

In calculating the values for the more dilute acid, in the third column of the previous table, none of the approximations shown in Example 3 were valid, and solution of a quadratic equation was necessary until after the equivalence point. In the postequivalence region, excess OH- predominates, and the simple calculation suffices.

As noted in the above figure, the initial pH values are higher and the equivalence point pH is lower for the more dilute solution (curve B). At intermediate volumes, however, the pH values differ only slightly because of the buffering action of the acetic acid/sodium acetate system. The change in OH- in the vicinity of the equivalence point becomes smaller with lower analyte and reagent concentrations. This effect is analogous to the effect with the titration of a strong acid with a strong base.

The Effect of Reaction Completeness

Titration curves for 0.1000 M solutions of acids with different dissociation constants are shown in the following figure. The pH change in the equivalence-point region becomes smaller as the acid becomes weaker-that is, as the reaction between the acid and the base becomes less complete.

Choosing an Indicator: The Feasibility of Titration

As noted from the previous figure, the choice of indicator is more limited for the titration of a weak acid than for that of a strong acid. For example, bromocresol green is totally unsuited for titration of 0.1000 M acetic acid. Bromothymol blue does not work either. An indicator exhibiting a color change in the basic region, such as phenolphthalein, however, should provide a sharp end point with a minimal titration error.

Titration Curves For Weak Bases

The calculations needed to draw the titration curve for a weak base are analogous to those for a weak acid.

Example 4

A 50.00-mL aliquot of 0.0500 M NaCN is titrated with 0.1000 M HCl. The reaction is:

CN + H3O+ ↔ HCN + H2O

Calculate the pH after the addition of (a) 0.00. (b) 10.00, (C) 25.00, and (d) 26.00 mL of acid.

(a) 0.00 mL of Reagent

The pH of a solution of NaCN can be calculated from Kb

CN-+ H2O ↔ HCN + OH-

Kb = [HCN][OH-]/[CN-] = x2/[CN-] = Kw/Ka = 1.00 X 10-14/6.20 X 10-10 = 1.6 X 10-5

[OH-] = x = (1.6 X 10-5 X 0.0500)1/2 = 8.97 X 10-4 M

pH = 14.00 - (- log 8.97 X 10-4) = 10.95

(b) 10.00 mL of Reagent: (Buffer)

[CN-] = (50.00 X 0.0500 - 10.00 X 0.1000)/60.00 = 1.500/60.00 M

[HCN] = (10.00 X 0.1000)/60.00 = 1.000/60.00 M

[H3O+] = Ka[HCN]/[CN-] = 6.2 X 10-10 X (1.000/60.00)/(1.500/60.00)

= 4.13 X 10-10 M

pH = - log (4.13 X 10-10) = 9.38

(c) 25.00 mL of Reagent

The solution contains the weak acid HCN, pH is calculated from Ka

[HCN] = (25.00 X 0.1000)/75.00 = 0.03333 M

[H3O+] = (Ka[HCN])1/2 = (6.2 X 10-10 X 0.03333)1/2 = 4.45 X 10-6 M

pH = - log (4.45 X 10-6) = 5.34

(d) 26.00 mL of Reagent: (Excess acid)

[H3O+] = [HCl] = (26.00 X 0.1000 - 50.00 X 0.0500)/76.00 = 1.32 X 10-3 M

pH = - log (l.32 X 10-3) = 2.88

The following figure shows hypothetical titration curves for a series of weak bases of different strengths. The curves show that indicators with acidic transition ranges must be used for weak bases.