Question

Two weak bases $AOH$ and $BOH$ having equal concentrations of $0.5M$ 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 $1.7\times {10}^{-9}$ respectively.(Report the answer upto second decimal)

Answer 1

For AOH :

$AOH\left(aq\right)\rightleftharpoons O{H}^{-}\left(aq\right)+A^{+}\left(aq\right)$

$att=0C_{1}00$

$att=t_{eq}{C}_1-C_1{\alpha }_1(C_1{\alpha }_1+C_2{\alpha }_2)C_1{\alpha }_1$

For BOH :

$BOH\left(aq\right)\rightleftharpoons O{H}^{-}\left(aq\right)+B^{+}\left(aq\right)$

$att=0C_{2}00$

$att=t_{eq}{C}_2-C_2{\alpha }_2(C_1{\alpha }_1+C_2{\alpha }_2)C_2{\alpha }_2$

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 ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak bases.So $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.
$K_{b_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$

$K_{b_1}{C}_1=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)$
Similarly,
$K_{b_2}{C}_2=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)$
Now,
$\frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}\frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{\left(C_1{\alpha }_1\right)}{\left(C_2{\alpha }_2\right)}=\frac{\left[A^{+}\right]}{\left[B^{+}\right]}\frac{\left[A^{+}\right]}{\left[B^{+}\right]}=\frac{K_{b1}{C}_1}{K_{b2}{C}_2}=\frac{1.1\times {10}^{-8}\times 0.5}{1.7\times {10}^{-9}\times 0.5}\frac{\left[A^{+}\right]}{\left[B^{+}\right]}=6.47$