Question

Two weak bases $AOH$ and $BOH$ having equal concentrations of 1 M are present in solution at ${25}^{o}C$ having dissociation constants as $1\times {10}^{-10}$ and $3\times {10}^{-10}$ respectively. Calculate the ratio $\frac{{\alpha }_1}{{\alpha }_2}$

Here ${\alpha }_1$ is the degree of dissociation of $AOH$
${\alpha }_2$ is degree of dissociation of $BOH$

• A. 1.5
• B. 1
• C. 0.5
• D. 0.33

Answer 1

The correct option is D 0.33

$C_1=C_2=1M{K}_{b1}=1\times {10}^{-10}M{K}_{b2}=3\times {10}^{-10}M$
For two bases :

Weak Bases $W{B}_1+W{B}_2$ :
Mixture of two weak bases $BOH$ and $AOH$:

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 $K_{b_2}$ :
$K_{b_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$

$C_1{K}_{b_1}+C_2{K}_{b_2}=(C_1{\alpha }_1+C_2{\alpha }_2{)}^2$
$\left[O{H}^{-}\right]=C_1{\alpha }_1+C_2{\alpha }_2=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}$
$\left[O{H}^{-}\right]=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}$
$\left[O{H}^{-}\right]=\sqrt{1(1\times {10}^{-10}+3\times {10}^{-10})}$
$\left[O{H}^{-}\right]=2\times {10}^{-5}M$
Since , both are weak acids, so ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak monoprotic acids. So, $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.

So:

$K_{b_1}\times C_1=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)$
$K_{b_2}\times C_2=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)$
Dividing both the above equations, we get:
$\frac{K_{b_1}}{K_{b_2}}=\frac{{\alpha }_1}{{\alpha }_2}$

$\frac{{\alpha }_1}{{\alpha }_2}=\frac{1\times {10}^{-10}}{3\times {10}^{-10}}\frac{{\alpha }_1}{{\alpha }_2}=0.33$