Question

Caculate pH of resultant solution of $0.1MHA+0.1MHB$
$\left[K_a\right(HA)=2\times {10}^{-5};K_a(HB)=4\times {10}^{-5}]$
log(2.44)=0.387
$\sqrt{6}=2.44$

• A. 2.61
• B. 4.56
• C. 3.49
• D. 1.02

Answer 1

The correct option is A 2.61

From the given $K_a$ values, it is concluded that these acids are weak acids.
We know, when mixture of weak acids are given
$\left[H^{+}\right]=\sqrt{K_{a_1}\times C_1+K_{a_2}\times C_2}$
$K_a\left(HA\right)=K_{a_1}=2\times {10}^{-5}$
$K_a\left(HB\right)=K_{a_2}=4\times {10}^{-5}$
Hence, $\left[H^{+}\right]=\sqrt{2\times {10}^{-5}\times 0.1+4\times {10}^{-5}\times 0.1}$
$\Rightarrow \left[H^{+}\right]=\sqrt{2\times {10}^{-6}+4\times {10}^{-6}}$
$\Rightarrow \left[H^{+}\right]=\sqrt{6\times {10}^{-6}}$
$\Rightarrow \left[H^{+}\right]=2.44\times {10}^{-3}$
So, $pH=-log\left[H^{+}\right]$
$\Rightarrow pH=-log(2.44\times {10}^{-3})$
$\Rightarrow pH=3-log\left(2.44\right)$
$\Rightarrow pH=3-0.387$
$\Rightarrow pH=2.61$