The dibasic compound B (pK_b1 = 4.00, pK_b2 = 8.00) was titrated with 1.00 M HCl. The initial solution of B was 0.100 M and had a volume of 100.0 mL. Find the pH at the following volumes of acid added and make a graph of pH versus V_a: V_a = 0, 1, 5, 9, 10, 11, 15, 19, 20, and 22 mL

The initial solution of B was 0.100 M and had a volume of 100.0 mL.

$\begin{array}{rcl}{K}_{b1}& =& \frac{\left[B{H}^{+}\right]\left[O{H}^{-}\right]}{\left[B\right]}\\ K_{b1}& =& {10}^{-4}\\ \frac{\left[B{H}^{+}\right]\left[O{H}^{-}\right]}{\left[B\right]}& =& {10}^{-4}\\ \frac{x.x}{\left[0.100-x\right]}& =& {10}^{-4}\\ x& =& 3.11\times {10}^{-3}M\end{array}$

Addition of 0 mL acid

$\begin{array}{rcl}pH& =& -\log \frac{K_w}{x}\\ & =& -\log \frac{1.00\times {10}^{-14}}{3.11\times {10}^{-3}}\\ & =& 11.49\end{array}$

Addition of 1 mL acid

$\begin{array}{rcl}pH& =& p{K}_{B{H}^{+}}+\log \frac{\left[B\right]}{\left[B{H}^{+}\right]}\\ & =& 10.00+\log \frac{9}{1}\\ & & 10.95\end{array}$

Addition of 10 mL acid

Concentration of BH⁺ will be

$\begin{array}{rcl}\frac{100}{110}\left(0.1\right)& =& 0.0909M\\ \left[H^{+}\right]& =& \sqrt{\frac{{10}^{-6}.{10}^{-10}\left(0.0909\right)+{10}^{-6}.{10}^{-14}}{{10}^{-6}+0.0909}}\\ & =& 1.00\times {10}^{-8}\\ pH& =& 8.00\end{array}$

Addition of 11 mL acid

localid="1660224059584" $\begin{array}{rcl}pH& =& p{K}_{B{H_2}^{+}}+\log \frac{\left[B{H}^{+}\right]}{\left[B{H_2}^{+}\right]}\\ & =& 6.00+\log \frac{9}{1}\\ & =& 6.95\end{array}$

Addition of 20 mL acid

localid="1660224384689" $\begin{array}{rcl}{K}_b& =& \frac{\left[B{H}^{+}\right]\left[H^{+}\right]}{\left[B{H_2}^{2+}\right]}\\ K_b& =& {10}^{-6}\\ \frac{\left[B{H}^{+}\right]\left[H^{+}\right]}{\left[B{H_2}^{2+}\right]}& =& {10}^{-6}\\ \frac{\left[x\right]\left[x\right]}{\left[0.0833\right]}& =& {10}^{-6}\\ x& =& 2.88\times {10}^{-4}M\\ \left[H^{+}\right]& =& 2.88\times {10}^{-4}M\\ pH& =& 3.54\end{array}$

Addition of 22 mL acid

localid="1660224177825" $\begin{array}{rcl}\left[H^{+}\right]& =& \frac{2}{122}\left(1.00\right)=1.64\times {10}^{-2}M\\ pH& =& 1.79\end{array}$

Following the above calculations, the pH of the solution is tabulated below

The graph is shown below