Question

Calculate $pH$ of a resultant solution of $0.1MHA(K_a={10}^{-6})$ and $0.45MHB(K_a=2\times {10}^{-6})$ at ${25}^{\circ }C$.

Answer 1

Given that,

$HA=0.1M(K_a={10}^{-6})$

$HB=0.45M(K_a=2\times {10}^{-6})$

We know that,

$H^{+}=\sqrt{K_a\times C}$

For $HA,$

$H^{+}=\sqrt{0.1\times {10}^{-6}}=0.316\times {10}^{-3}$

For $HB$,

$H^{+}=\sqrt{0.45\times 2\times {10}^{-6}}=0.948\times {10}^{-3}$

Resulting

$H^{+}=\frac{(0.316\times {10}^{-3}+0.948\times {10}^{-3})}{2}$

$H^{+}=\frac{1.265}{2}\times {10}^{-3}=0.632\times {10}^{-3}$

$pH=-\log (0.632\times {10}^{-3})$

$pH=2.8$