Question 19.4: Finding the pH During a Weak Acid–Strong Base Titration Prob......

Plan (a) 0.00 mL: No base has been added yet, so this is a weak-acid dissociation. We calculate the pH as we did in Section 18.4. (b) 30.00 mL: We find the amount (mol) of Pr^− and of HPr, and substitute into the Henderson-Hasselbalch equation to solve for pH. (c) 40.00 mL: The amount (mol) of NaOH added equals the initial amount (mol) of HPr, so a solution of Na^+ and Pr^− exists. We calculate the pH as we did in Section 18.6, except that we need total volume to find [Pr^−]. (d) 50.00 mL: We calculate the amount (mol) of excess OH^− in the total volume, convert to [H_3O^+] and then to pH.

Solution (a) 0.00 mL of 0.1000 M NaOH added. Following the approach used in Sample Problem 18.8 and just described in the text (Table 1), we obtain

       HPr(aq)  +  H_2O(l) \xrightleftharpoons[]{}  H_3O^+(aq)  +  Pr^−(aq)              K_a  =  \frac{[H_3O^+][Pr^−]}{[HPr]}  =  1.3×10^{−5}
             K_a  =  \frac{[H_3O^+][Pr^−]}{[HPr]}  =  1.3×10^{−5}  ≈  \frac{(x)(x)}{0.1000}
            [H_3O^+]  =  x  ≈  \sqrt{K_a  ×  [HPr]_{\text{init}}}  =  \sqrt{(1.3×10^{−5})(0.1000)}  =  1.1×10^{−3}  M
            pH  =  2.96

(b) 30.00 mL of 0.1000 M NaOH added. Calculating the initial amounts of HPr and OH^−:

For every mole of NaOH, 1 mol of Pr^− forms, so we have this stoichiometry reaction table (Table 2):

The last line of the table gives the new initial amounts of HPr and Pr^− that react to attain a new equilibrium. With x very small, we assume that the [HPr]/[Pr^−] ratio at equilibrium is essentially equal to the ratio of these new initial amounts (see Comment in Sample Problem 19.1).

Calculating pK_a:
             pK_a  =  −\log  K_a  =  −\log  (1.3×10^{−5})  =  4.89
Solving for pH using the Henderson-Hasselbalch equation:
             pH  =  pK_a  +  \log  \frac{[Pr^−]}{[HPr]}  =  4.89  +  \log  \frac{0.003000 }{0.001000}
             pH  =  5.37

(c) 40.00 mL of 0.1000 M NaOH added. Calculating the initial amounts of HPr and OH^−:

We have this stoichiometry reaction table (Table 3):

Calculating [Pr^−] after all HPr has reacted:
             [Pr^−]  =  \frac{0.004000\text{ mol}}{0.04000  L  +  0.04000  L}  =  0.05000  M
Setting up a reaction table for the base dissociation (Table 4), with x  =  [Pr^−]_{\text{reacting}}  =  [OH^−] and assuming [Pr^−]_{\text{reacting}}  <<  [Pr^−]_{\text{init}}, calculating K_b, and solving for x (see Sample Problem 18.12 for similar steps):

            K_b  =  \frac{K_w}{K_a}  =  \frac{1.0×10^{−14}}{1.3×10^{−5}}  =  7.7×10^{−10}
            K_b  =  \frac{[HPr][OH^−]}{[Pr^−]}  =  7.7×10^{−10}  ≈  \frac{(x)(x)}{0.05000}
            [OH^−]  =  x  ≈  \sqrt{K_b  ×  [Pr^−]_{\text{init}}}  =  \sqrt{(7.7×10^{−10})(0.05000)}  =  6.2×10^{−6}  M
            [H_3O^+]  =  \frac{K_w }{[OH^−]}  =  \frac{1.0×10^{−14} }{6.2×10^{−6}}  =  1.6×10^{−9}  M
            pH  =  8.80

(d) 50.00 mL of 0.1000 M NaOH added. Calculating the initial amounts of HPr and OH^−:

We have this stoichiometry reaction table (Table 5):

            [OH^−]  =  \frac{\text{amount (mol) of excess }OH^−}{\text{total volume}}  =  \frac{\text{0.001000 mol}}{0.04000  L  +  0.05000  L}  =  0.01111  M
            [H_3O^+]  =  \frac{K_w}{[OH^−]}  =  \frac{1.0×10^{−14}}{0.01111}  =  9.0×10^{−13}  M
            pH  =  12.05

Check As expected, the pH increases through the four regions of the titration. Be sure to round off and check the arithmetic along the way.