Question

The dissociation constants of two weak bases $AOH$ and $BOH$ are $K_{b_1}=1\times {10}^{-8}$ and $K_{b_2}=2\times {10}^{-9}$ respectively. The $H^{+}$ ions concentration of the solution that is simultaneously 0.01 M in $AOH$ and 0.05 M in $BOH$ at ${25}^{\circ }C$ is:

• A. $1.41\times {10}^{-5}M$
• B. $7.07\times {10}^{-10}M$
• C. $9.5\times {10}^{-12}M$
• D. $4\times {10}^{-6}M$

Answer 1

The correct option is B $7.07\times {10}^{-10}M$

$K_{b1}=1\times {10}^{-8}M$
$K_{b2}=2\times {10}^{-9}M$
$C_1=0.01M{C}_2=0.05M$
$\left[O{H}^{-}\right]=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}\left[O{H}^{-}\right]=\sqrt{(0.01\times 1\times {10}^{-8})+(0.05\times 2\times {10}^{-9})}$
$\left[O{H}^{-}\right]=\sqrt{2\times {10}^{-10}}\left[O{H}^{-}\right]=1.41\times {10}^{-5}M$
At ${25}^{\circ }C$ ,
$K_w={10}^{-14}=\left[H^{+}\right]\left[O{H}^{-}\right]\left[H^{+}\right]=\frac{{10}^{-14}}{\left[O{H}^{-}\right]}=\frac{{10}^{-14}}{1.41\times {10}^{-5}}\left[H^{+}\right]=7.07\times {10}^{-10}M$


Theory:

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 $K_{b_2}$ :
$K_{b_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$
Since ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak bases.So $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.
$C_1{K}_{b_1}+C_2{K}_{b_2}=(C_1{\alpha }_1+C_2{\alpha }_2{)}^2$
$\left[O{H}^{-}\right]=C_1{\alpha }_1+C_2{\alpha }_2=\sqrt{C_1{K}_{b_1}+C_2{K}_{b_2}}$