Question

Calculate the pH of 1 M $C{H}_{3}COONa$ solution. $(K_a=1.8\times {10}^{-5})$
Take $log\left(1.8\right)=0.26$

• A. 9.37
• B. 10.46
• C. 8.52
• D. 11.49

Answer 1

The correct option is A 9.37

Salt of a weak acid and strong base:
$pH=7+\frac{1}{2}(p{K}_a+logC)K_a=1.8\times {10}^{-5}p{K}_a=-log\left(K_a\right)p{K}_a=-log(1.8\times {10}^{-5})=(5-0.26)=4.74pH=7+\frac{1}{2}(4.74+log(1\left)\right)pH=9.37$



Theory:
Salt of weak acid $WA$ and a strong base $SB$
Consider, weak acid ($C{H}_{3}COOH$) and strong base ($NaOH$) together form a Salt of ($C{H}_{2}COONa$)
On complete dissociation $C{H}_{3}COONa\left(aq\right)\to C{H}_{3}CO{O}^{-}\left(aq\right)+N{a}^{+}\left(aq\right)$
On hydrolysis $C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right))$
Note:- Only the anion will undergo hydrolysis and the solution will be alkaline in nature due to $O{H}^{-}$ formation.
So here, the anion is considered responsible for the alkaline nature of the solution.
$\Rightarrow pH>7$
Consider a salt $\left(MX\right)$ added to water and produce cations $\left(M^{+}\right)$and anions $\left(X^{-}\right)$ in water. After dissociation $\left(O{H}^{-}\right)$ concentration increases.
For example :- $C{H}_{3}CO{O}^{-}\left(aq\right)+H_{2}O\left(l\right)\rightleftharpoons C{H}_{3}COOH\left(aq\right)+O{H}^{-}\left(aq\right)$
$C00$
$C-ChChCh$
In this reaction the $\left(K_h\right)$ hydrolysis constant is:
$K_h=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}=\frac{(Ch{)}^2}{C(1-h)}$
$C{H}_{3}COOH\left(aq\right)\rightleftharpoons C{H}_{3}CO{O}^{-}\left(aq\right)+O{H}^{-}\left(aq\right)$
$K_a=\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}$
$H_{2}O\left(l\right)\rightleftharpoons H^{+}\left(aq\right)+O{H}^{-}\left(aq\right)$
In this reaction the $K_w$ constant is:
$K_w=\left[H^{+}\right]\left[O{H}^{-}\right]$
$\frac{K_w}{K_a}=\frac{\left[H^{+}\right]\left[O{H}^{-}\right]}{\frac{\left[C{H}_{3}CO{O}^{-}\right]\left[H^{+}\right]}{\left[C{H}_{3}COOH\right]}}$
$\frac{K_w}{K_a}=\frac{\left[C{H}_{3}COOH\right]\left[O{H}^{-}\right]}{\left[C{H}_{3}CO{O}^{-}\right]}$
$K_h=\frac{K_w}{K_a}$
Form earlier equation of $K_h$
$K_h=\frac{(Ch{)}^2}{C(1-h)}$ Generally h<<1. So, neglect(1-h)
If h<<1
$C{h}^2=K_h=\frac{K_w}{K_a}$
Then $h=\sqrt{\frac{K_h}{C}}=\sqrt{\frac{K_w}{K_{a}C}}$
$\left[O{H}^{-}\right]=C\times \sqrt{\frac{K_w}{K_{a}C}}=\sqrt{\frac{K_{w}C}{K_a}}$
$\left[H^{+}\right]=\frac{K_w}{\left[O{H}^{-}\right]}=\sqrt{\frac{K_w{K}_a}{C}}$
$pH=\frac{1}{2}[p{K}_w+p{K}_a+logC]$.
Only valid if $h<0.1$ or 10%
$p{K}_a=-log{K}_a$
$pH=7+\frac{1}{2}[p{K}_a+logC]$ at ${25}^{\circ }$