Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\begin{array}{cccccccc}& \text{BOH}& +& \text{HCl}& ⟶& \text{BCl}& +& {\text{H}}_2\text{O}\\ \text{Initial}& 1\text{.}6\text{mM}& & 1\text{.}6\text{mM}& & 0& & 0\\ & 0& & 0& & 1\text{.}6\text{mM}& & 1\text{.}6\text{mM}\end{array}$

Total volume = 40 mL

[BCl] = 1.6/40 = 0.04M

$\text{p}\text{H}=\frac{1}{2}\left[\text{p}{\text{K}}_{\text{w}}-\text{p}{\text{K}}_{\text{b}}-\text{log}\text{C}\right]$

$5=\frac{1}{2}\left[14-\text{p}{\text{K}}_{\text{b}}-\text{log }0\text{.}04\right]$

pK_b= 5.4

$\begin{array}{ccccccc}\text{BCl}& +& \text{NaOH}& ⟶& \text{BOH}& +& \text{NaCl}\\ 1\text{.}6\text{mM}& & 0\text{.}4\text{mM}& & 0& & 0\\ -0\text{.}4\text{mM}& & -0\text{.}4\text{mM}& & +0\text{.}4\text{mM}& & +0·4\text{mM}\\ \text{---------------}& & \text{---------------}& & \text{---------------}& & \text{---------------}\\ 1·2\text{mM}& & 0& & 0\text{.}4\text{mM}& & 0\text{.}4\text{mM}\end{array}$

$\text{p}\text{OH}=\text{p}{\text{K}}_{\text{b}}+\text{log}\frac{\left[{\text{B}}^{+}\right]}{\left[\text{BOH}\right]}$


$=5.4+\log \frac{1.2}{0.4}$

=5.4+log3

$=5\text{.}4+0\text{.}48$

$=5\text{.}88$