Question

Calculate the degree of dissociation $\left({\alpha }_1\right)$, in a solution obtained by mixing equal volume of 0.02 M AOH and 0.2 M BOH solution.
Where, ${\alpha }_1$ is the degree of dissociation of weak base AOH and ${\alpha }_2$ is the degree of dissociation of weak base BOH.
Given that,
$K_b\left(AOH\right)=1\times {10}^{-8}M$
$K_b\left(BOH\right)=2\times {10}^{-9}M$

• A. $7.8\times {10}^{-6}$
• B. $5.78\times {10}^{-4}$
• C. $4.55\times {10}^{-2}$
• D. $2.3\times {10}^{-3}$

Answer 1

The correct option is B $5.78\times {10}^{-4}$

Denoting a for AOH and b for BOH.
Let the equal volumes of solution a and b as V, i.e. $V_a=V_b=V$
So, $C_1=\frac{C_a{V}_a}{V_a+V_b}=\frac{0.02\times V}{V+V}{C}_1=0.01M$
Similarly, $C_2=\frac{C_b\times V_b}{V_a+V_b}=\frac{0.2\times V}{V+V}{C}_2=0.1M$
$K_{b1}=1\times {10}^{-8}M$
$K_{b2}=2\times {10}^{-9}M$

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 $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.
$C_1{K}_{b_1}+C_2{K}_{b_2}=(C_1{\alpha }_1+C_2{\alpha }_2{)}^2$
$\left[O{H}^{-}\right]=C_1{\alpha }_1+C_2{\alpha }_2=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}$
$\left[O{H}^{-}\right]=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}[O{H}^{-}{]}^{=}\sqrt{(0.01\times 1\times {10}^{-8})+(0.1\times 2\times {10}^{-9})}$
$\left[O{H}^{-}\right]=\sqrt{3\times {10}^{-10}}\left[O{H}^{-}\right]=1.73\times {10}^{-5}M$
From the equation of $K_{b1}$ and using approximation $1-{\alpha }_1\approx 1$
$K_{b1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1}{K}_{b1}=\left[O{H}^{-}\right]\left({\alpha }_1\right)$

$\because \left[O{H}^{-}\right]=(C_1{\alpha }_1+C_2{\alpha }_2)$
${\alpha }_1=\frac{K_{b1}}{\left[O{H}^{-}\right]}=\frac{1\times {10}^{-8}}{1.73\times {10}^{-5}}=5.78\times {10}^{-4}$