Problem 62
Sketch the titration curve for the titration of a generic weak base \(\mathrm{B}\) with a strong acid. The titration reaction is $$ \mathrm{B}+\mathrm{H}^{+} \rightleftharpoons \mathrm{BH}^{+} $$ On this curve, indicate the points that correspond to the following: a. the stoichiometric (equivalence) point b. the region with maximum buffering c. \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}}\) d. \(\mathrm{pH}\) depends only on \([\mathrm{B}]\) e. \(\mathrm{pH}\) depends only on \(\left[\mathrm{BH}^{+}\right]\) f. \(\mathrm{pH}\) depends only on the amount of excess strong acid added
Problem 63
Consider the titration of 40.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M} \mathrm{HClO}_{4}\) by 0.100 \(\mathrm{M}\) KOH. Calculate the pH of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 80.0 \mathrm{mL}} \\ {\text { b. } 10.0 \mathrm{mL}} & {\text { e. } 100.0 \mathrm{mL}} \\ {\text { c. } 40.0 \mathrm{mL}}\end{array} $$
Problem 64
Consider the titration of 80.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{Ba}(\mathrm{OH})_{2}\) by 0.400 \(\mathrm{M} \mathrm{HCl}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of HCl have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}} \\ {\text { c. } 30.0 \mathrm{mL}} & {\text { e. } 80.0 \mathrm{mL}}\end{array} $$
Problem 65
Consider the titration of 100.0 \(\mathrm{mL}\) of 0.200 \(\mathrm{M}\) acetic acid \(\left(K_{\mathrm{a}}=1.8 \times 10^{-5}\right)\) by 0.100 \(\mathrm{M} \mathrm{KOH}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of KOH have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 150.0 \mathrm{mL}} \\ {\text { b. } 50.0 \mathrm{mL}} & {\text { e. } 200.0 \mathrm{mL}} \\ {\text { c. } 100.0 \mathrm{mL}} & {\text { f. } 250.0 \mathrm{mL}}\end{array} $$
Problem 66
Consider the titration of 100.0 \(\mathrm{mL}\) of 0.100 \(\mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) \(\left(K_{\mathrm{b}}=3.0 \times 10^{-6}\right)\) by 0.200\(M \mathrm{HNO}_{3}\) . Calculate the \(\mathrm{pH}\) of the resulting solution after the following volumes of \(\mathrm{HNO}_{3}\) have been added. $$ \begin{array}{ll}{\text { a. } 0.0 \mathrm{mL}} & {\text { d. } 40.0 \mathrm{mL}} \\ {\text { b. } 20.0 \mathrm{mL}} & {\text { e. } 50.0 \mathrm{mL}} \\ {\text { c. } 25.0 \mathrm{mL}} & {\text { f. } 100.0 \mathrm{mL}}\end{array} $$
Problem 67
Lactic acid is a common by-product of cellular respiration and is often said to cause the "burn" associated with strenuous activity. A 25.0 -mL sample of 0.100\(M\) lactic actid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{3}\right.\) \(\mathrm{p} K_{\mathrm{a}}=3.86 )\) is titrated with 0.100 \(\mathrm{M}\) NaOH solution. Calculate the pH after the addition of \(0.0 \mathrm{mL}, 4.0 \mathrm{mL}, 8.0 \mathrm{mL}, 12.5 \mathrm{mL},\) \(20.0 \mathrm{mL}, 24.0 \mathrm{mL}, 24.5 \mathrm{mL}, 24.9 \mathrm{mL}, 25.0 \mathrm{mL}, 25.1 \mathrm{mL}\) \(26.0 \mathrm{mL}, 28.0 \mathrm{mL}\) , and 30.0 \(\mathrm{mL}\) of the NaOH. Plot the results of your calculations as \(\mathrm{pH}\) versus milliliters of NaOH added.
Problem 68
Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) propanoic acid \(\left(\mathrm{HC}_{3} \mathrm{H}_{3} \mathrm{O}_{2}, K_{\mathrm{a}}=1.3 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M} \mathrm{NaOH}\) .
Problem 69
Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M \mathrm{NH}_{3}\left(K_{\mathrm{b}}=1.8 \times 10^{-5}\right)\) with 0.100 \(\mathrm{M}\) \(\mathrm{HCl} .\)
Problem 70
Repeat the procedure in Exercise \(67,\) but for the titration of 25.0 \(\mathrm{mL}\) of 0.100\(M\) pyridine with 0.100\(M\) hydrochloric acid \(\left(K_{\mathrm{b}} \text { for pyridine is } 1.7 \times 10^{-9}\right) .\) Do not calculate the points at 24.9 and 25.1 \(\mathrm{mL}\)
Problem 71
Calculate the \(\mathrm{pH}\) at the halfway point and at the equivalence point for each of the following titrations. $$ \begin{array}{l}{\text { a. } 100.0 \mathrm{mL} \text { of } 0.10 \mathrm{M} \mathrm{HC}_{7} \mathrm{HsO}_{2}\left(K_{\mathrm{a}}=6.4 \times 10^{-5}\right) \text { titrated }} \\ {\text { by } 0.10 \mathrm{M} \mathrm{NaOH}}\end{array} $$ $$ \begin{array}{l}{\text { b. } 100.0 \mathrm{mL} \text { of } 0.10 M \mathrm{C}_{2} \mathrm{H}_{5} \mathrm{NH}_{2}\left(K_{\mathrm{b}}=5.6 \times 10^{-4}\right) \text { titrated }} \\ {\text { by } 0.20 \mathrm{M} \mathrm{HNO}_{3}}\end{array} $$ $$ 100.0 \mathrm{mL} \text { of } 0.50 \mathrm{M} \mathrm{HCl} \text { titrated by } 0.25 \mathrm{M} \mathrm{NaOH} $$
