Question

Two weak monobasic acids HA and HB having equal concentrations of 0.5 M are present in solution at ${25}^{o}C$ having dissociation constants as $3.5\times {10}^{-10}$ and $4.5\times {10}^{-10}$ respectively. Calculate the pH of the solution at given temperature.

• A. 4.7
• B. 5.7
• C. 6.7
• D. 2.5

Answer 1

The correct option is A 4.7

$C_1=C_2=0.5M{K}_{a1}=3.5\times {10}^{-10}M{K}_{a2}=4.5\times {10}^{-10}M$
For two weak acids :

$\left[H^{+}\right]=\sqrt{C_1{K}_{a_1}+C_2{K}_{a_2}}$
$\left[H^{+}\right]=\sqrt{0.5(3.5\times {10}^{-10}+4.5\times {10}^{-10})}$
$\left[H^{+}\right]=2\times {10}^{-5}MpH=-log\left[H^{+}\right]pH=-log(2\times {10}^{-5})=(5-log2)pH=4.7$
$\therefore$ The pH of HA and HB in solution is 4.7

Theory :
Weak acid 1 $\left(W{A}_1\right)$ + Weak acid 2 $\left(W{A}_2\right)$ :

Mixture of two weak monoprotic acids HA and HB with concentration $C_{1}and{C}_2$ and degree of dissociation ${\alpha }_{1}and{\alpha }_2$

For HA :

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

$att=0C_{1}00$

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

For HB :

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

$att=0C_{2}00$

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

Dissociation constant for HA $K_{a_1}$ :

$K_{a_1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$

Dissociation constant for HB $K_{a_2}$ :

$K_{a_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}{C_2(1-{\alpha }_2)}$

Since ${\alpha }_{1}and{\alpha }_2$ are very small in comparison to unity for weak monoprotic acids.So $1-{\alpha }_1\approx 1and1-{\alpha }_2\approx 1$.

$C_1{K}_{a_1}+C_2{K}_{a_2}=(C_1{\alpha }_1+C_2{\alpha }_2{)}^2$

$\left[H^{+}\right]=C_1{\alpha }_1+C_2{\alpha }_2=\sqrt{C_1{K}_{a_1}+C_2{K}_{a_2}}$