Question

The dissociation constants of two weak acids HA and HB are $3\times {10}^{-4}M$ and $5\times {10}^{-4}M$ respectively. The equilibrium ratio of $A^{-}$ to $B^{-}$ in a solution that is simultaneously 0.5 M in HA and 0.75 M in HB is :

• A. 0.4
• B. 1.5
• C. 2.5
• D. 0.75

Answer 1

The correct option is A 0.4

$HA(aq.)\rightleftharpoons H^{+}(aq.)+A^{-}(aq.)Initial:C_{1}00Equilibrium:C_1(1-{\alpha }_1)C_1{\alpha }_1+C_2{\alpha }_2{C}_1{\alpha }_1$

$HB(aq.)\rightleftharpoons H^{+}(aq.)+B^{-}(aq.)Initial:C_{2}00Equilibrium:C_2(1-{\alpha }_2)C_1{\alpha }_1+C_2{\alpha }_2{C}_2{\alpha }_2$

$K_{a1}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{C_1(1-{\alpha }_1)}$

${\alpha }_1<<1\therefore 1-{\alpha }_1\approx 1K_{a1}{C}_1=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)$
Similarly,
$K_{a2}{C}_2=(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)$
Now,
$\frac{K_{a1}{C}_1}{K_{a2}{C}_2}=\frac{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_1{\alpha }_1\right)}{(C_1{\alpha }_1+C_2{\alpha }_2)\left(C_2{\alpha }_2\right)}\frac{K_{a1}{C}_1}{K_{a2}{C}_2}=\frac{\left(C_1{\alpha }_1\right)}{\left(C_2{\alpha }_2\right)}=\frac{\left[A^{-}\right]}{\left[B^{-}\right]}\frac{\left[A^{-}\right]}{\left[B^{-}\right]}=\frac{K_{a1}{C}_1}{K_{a2}{C}_2}=\frac{3\times {10}^{-4}\times 0.5}{5\times {10}^{-4}\times 0.75}\frac{\left[A^{-}\right]}{\left[B^{-}\right]}=0.4$