Question

0.5 M HA and 0.4 M HB are present in two different solutions. The volume of solution containing HA is 10 L whereas volume of solution containing HB is 30 L . The dissociation constants of both the weak acids HA and HB are $3\times {10}^{-4}$ and $5\times {10}^{-4}$ respectively.
The pH of solution at equilibrium after both HA and HB solutions are mixed together is :
given: $lo{g}_{10}\left(1.875\right)=0.27$

• A. 8.46
• B. 1.86
• C. 4.98
• D. 6.05

Answer 1

The correct option is B 1.86

$C_a=0.5M{C}_b=0.4M{V}_a=10L{V}_b=30L$

Concentration of HA after mixing= $\frac{C_a{V}_a}{V_a+V_b}=\frac{0.5\times 10}{10+30}=0.125M=C_1$

Concentration of HB after mixing= $\frac{C_b{V}_b}{V_a+V_b}=\frac{0.4\times 30}{10+30}=0.3M=C_2$

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

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

$pH=-\frac{1}{2}log\left[\right(K_{a1}{C}_1)+(K_{a2}{C}_2\left)\right]pH=-\frac{1}{2}log\left[\right(3\times {10}^{-4}\times 0.125)+(5\times {10}^{-4}\times 0.3\left)\right]pH=-\frac{1}{2}log(1.875\times {10}^{-4})pH=\frac{1}{2}(4-log1.875)=\frac{1}{2}(4-0.27)pH=1.865$