Question

$100ml$ of $0.2MNaOH$ are mixed with $100ml$ of $0.2MC{H}_{3}COOH$, the pH of the resulting solution would be nearly:

• A. $1$
• B. $0.7$
• C. $8.875$
• D. $1.6$

Answer 1

The correct option is C $8.875$

$C{H}_{3}COOH$ is a weak acid & $NaOH$ is a strong base

$C{H}_{3}COOH+NaOH⟶C{H}_{3}COONa$

$20mmol$ $20mmol$ $20mmol$ $K_a=1.8\times {10}^{-5}$

$H_{2}O\rightleftharpoons H^{+}{+}^{-}OH$ $K_w$

$C{H}_{3}CO{O}^{-}+H^{+}\rightleftharpoons C{H}_{3}COOH$ $K^{\prime }=1/K_a$

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH$

$K_h=K_w{K}^{\prime }=\frac{K_w}{K_a}=K_b$ as $K_w=K_a\times K_b$

$C{H}_{3}CO{O}^{-}+H_{2}O\rightleftharpoons C{H}_{3}COOH{+}^{-}OH$

$C$ - -

$C-Ch$ $Ch$ $Ch$

$K_h=\frac{C^2{h}^2}{C(1-h)}$ as $h<<1$ thus $(1-h)\simeq 1$

$\Rightarrow K_h=c{h}^2$ or $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_{a}C}}$

$\Rightarrow {[}^{-}OH]=Ch=C\sqrt{\frac{K_w}{K_{a}C}}=\sqrt{\frac{K_{w}C}{K_a}}$

or $pOH=\frac{1}{2}[p{K}_w-p{K}_a-\log C]$

$\Rightarrow pOH=\frac{1}{2}[14-4.74-\log \frac{20mmol}{200ml}]$

$\Rightarrow pOH=\frac{1}{2}[14-4.75-(-1\left)\right]$

$\Rightarrow pOH=5.128$

$\Rightarrow pH=14-pOH=14-5.128$

$\Rightarrow pH=8.875$