Question

The pH of the solution $0.1M$ $HA+0.1M$ $HB$ is:

[$K_{a\left(HA\right)}=2\times {10}^{-5};K_{a\left(HB\right)}=4\times {10}^{-5}]$

• A. $2.61$
• B. $7.4$
• C. $9.5$
• D. $11.39$

Answer 1

The correct option is A $2.61$

$\underset{0.1M}{HA}\rightleftharpoons \underset{a+bM}{H^{+}}+\underset{aM}{A^{-}}$

$K_a=\frac{(a+b)a}{0.1}=2\times {10}^{-5}$.....(1)

$\underset{0.1M}{HB}\rightleftharpoons \underset{a+bM}{H^{+}}+\underset{bM}{B^{-}}$

$K_a=\frac{(a+b)b}{0.1}=4\times {10}^{-5}$......(2)

Divide equation (1) by equation (2)

$\frac{(a+b)a}{(a+b)b}=\frac{2\times {10}^{-5}}{4\times {10}^{-5}}$

$\frac{a}{b}=0.5$

$a=0.5b$

Substitute this in equation (1)

$K_a=\frac{(0.5b+b)0.5b}{0.1}=2\times {10}^{-5}$

$0.75b^2=2\times {10}^{-6}$

$b=1.6329\times {10}^{-3}M$

$a=0.5b=0.5\times 1.6329\times {10}^{-3}M=8.165\times {10}^{-4}M$

$\left[H^{+}\right]=a+b=8.165\times {10}^{-4}M+1.6329\times {10}^{-3}M$

$\left[H^{+}\right]=2.449\times {10}^{-3}M$

$pH=-log\left[H^{+}\right]=-log(2.449\times {10}^{-3})M=2.61$