Question:Titrating weak acid with weak base.

(a) Prepare a family of graphs for the titration of 50.0 mL of 0.020 0 M HA (pK_a = 4.00) with 0.100 M B (pK_b = 3.00, 6.00, and 9.00).

(b) Write the acid-base reaction that occurs when acetic acid and sodium benzoate (the salt of benzoic acid) are mixed, and find the equilibrium constant for the reaction. Find the pH of a solution prepared by mixing 212 mL of 0.200 M acetic acid with 325 mL of 0.050 0 M sodium benzoate.

For acetic acid

$$\begin{gathered} K_a=1.75\times {10}^{-5} \\ {V}_a=212mL \\ {C}_a=0.2M \end{gathered}$$

For sodium benzoate

$$\begin{gathered} K_b=1.59\times {10}^{-10} \\ {V}_b=325mL \\ {C}_b=0.05M \end{gathered}$$

Fromtable 11.5 we can obtain the equation for the fraction of titration of weak acid (HA) with weak base (B)

$\varphi =\frac{C_b{V}_b}{C_a{V}_a}=\frac{{\alpha }_{A^{-}}+{\displaystyle \frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_a}}}{{\alpha }_{B{H}^{+}}-\frac{\left[H^{+}\right]-\left[O{H}^{-}\right]}{C_b}}$

Other formulas need to be used

$\begin{array}{rcl}{K}_{B{H}^{+}}& =& \frac{K_w}{K_b}\\ \left[H^{+}\right]& =& {10}^{-pH}\\ \left[O{H}^{-}\right]& =& \frac{K_w}{\left[H^{+}\right]}\\ \left[{\alpha }_{A^{-}}\right]& =& \frac{K_a}{\left[H^{+}\right]+K_a}\\ \left[{\alpha }_{B{H}^{+}}\right]& =& \frac{\left[H^{+}\right]}{\left[H^{+}\right]+K_{B{H}^{+}}}\end{array}$

In order to determine the equilibrium constant, the following can be written

The reaction between acetic acid and sodium benzoate is depicted below

$\begin{array}{rcl}C{H}_{3}COOH+C_6{H}_{5}COONa& \rightleftharpoons & C{H}_{3}COONa+C_6{H}_{5}COOH\\ Aceticacid+Sodiumbenzoate& \rightleftharpoons & Sodiumacetate+Benzoicacid\end{array}$

The value of the equilibrium constant(K) will be

role="math" localid="1662036119477" $\begin{array}{rcl}K& =& \frac{K_a}{K_{B{H}^{+}}}\\ K& =& \frac{K_a}{{\displaystyle \raisebox{1ex}{$K_w$}\!\left/ \!\raisebox{-1ex}{$K_B$}\right.}}\\ K& =& \frac{\left(1.75\times {10}^{-5}\right)\times \left(1.59\times {10}^{-10}\right)}{{\displaystyle {10}^{-14}}}\\ & =& 0.28\\ & & \end{array}$

Therefore, the equilibrium constant for the reaction is 0.28