Question

Two weak bases $AOH$ and $BOH$ having equal concentrations of $0.5\text{M}$ are present in a solution. Calculate the ratio, $\frac{\left[A^{+}\right]}{\left[B^{+}\right]}$ at equilibrium.
Dissociation constants for $AOH$ and $BOH$ are $1.1\times {10}^{-8}$ and $1.65\times {10}^{-9}$ respectively.

• A. $3.53$
• B. $0.067$
• C. $6.67$
• D. $0.035$

Answer 1

The correct option is C $6.67$

For $AOH$:
$\begin{array}{ccccc}& AOH\left(aq\right)& \rightleftharpoons & O{H}^{-}\left(\text{aq}\right)& +A^{+}\left(aq\right)\\ \text{at, t=0}& C_1& & 0& 0\\ \text{at, t}=t_{\text{equi}}& C_1-C_1{\alpha }_1& & C_1{\alpha }_1+C_2{\alpha }_2& C_1{\alpha }_1\end{array}$

For $BOH$:
$\begin{array}{ccccc}& BOH\left(aq\right)& \rightleftharpoons & O{H}^{-}\left(\text{aq}\right)& +B^{+}\left(aq\right)\\ \text{at, t=0}& C_2& & 0& 0\\ \text{at, t}=t_{\text{equi}}& C_2-C_2{\alpha }_2& & C_1{\alpha }_1+C_2{\alpha }_2& C_2{\alpha }_2\end{array}$

Dissociation constant for $AOH$, $K_{b_1}$ :

$K_{b_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$
Dissociation constant for $BOH$, $K_{b_2}$ :
$K_{b_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$
Since ${\alpha }_1\text{and}{\alpha }_2$ are very small in comparison to unity for weak bases, so $1-{\alpha }_1\approx 1\text{and}1-{\alpha }_2\approx 1$.
$\therefore K_{b_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1}$
$\Rightarrow K_{b_1}{C}_1=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)$
Similarly,
$K_{b_2}{C}_2=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)$
Now,
$\frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}\Rightarrow \frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{\left(C_1{\alpha }_1\right)}{\left(C_2{\alpha }_2\right)}=\frac{\left[A^{+}\right]}{\left[B^{+}\right]}\Rightarrow \frac{\left[A^{+}\right]}{\left[B^{+}\right]}=\frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{1.1\times {10}^{-8}\times 0.5}{1.65\times {10}^{-9}\times 0.5}\Rightarrow \frac{\left[A^{+}\right]}{\left[B^{+}\right]}=6.67$