Question

Trimethylamine, $(C{H}_3{)}_{3}N$ is a weak base $K_b=6.25\times {10}^{-5}$ that hydrolyzes by the following equilibrium reaction at room temperature:
$(C{H}_3{)}_{3}N+H_{2}O\rightleftharpoons (C{H}_3{)}_{3}N{H}^{+}+O{H}^{-}$

The pH of 0.1 M solution of $(C{H}_3{)}_{3}N{H}^{+}$ is:

• A. 5.4
• B. 6.4
• C. 7.2
• D. 4.9

Answer 1

The correct option is A 5.4

$\left(C{H}_3{)}_{3}N{H}^{+}\right(aq)+H_{2}O(l)\rightleftharpoons (C{H}_3{)}_{3}N\left(aq\right)+H_3{O}^{+}\left(aq\right)Initially:C00Equilibrium:C-C\alpha C\alpha C\alpha$
At, equilibrium, $\left[H_3{O}^{+}\right]=C\alpha$

We know that , for an aqueous solution at room temperature ,
$K_a\times K_b=K_{w}Here,K_w={10}^{-14}$
$K_w$ is Ionisation Constant for $H_{2}O$
$K_a=\frac{K_w}{K_b}{K}_a=\frac{{10}^{-14}}{6.25\times {10}^{-5}}=1.6\times {10}^{-10}$
By definition, the dissociation constant will be :
$K_a=\frac{\left[\right(C{H}_3{)}_{3}N]\times [H_3{O}^{+}]}{\left[\right(C{H}_3{)}_{3}N{H}^{+}]}=\frac{C\alpha \times C\alpha }{C-C\alpha }$

$K_a=\frac{(C\alpha {)}^2}{C(1-\alpha )}$
Since $\alpha$ is very less
so, $(1-\alpha \approx 1)$
$K_a\times C=(C\alpha {)}^2{K}_a\times C=[H_3{O}^{+}{]}^2\left[H_3{O}^{+}\right]=\sqrt{K_a\times C}=\sqrt{1.6\times {10}^{-10}\times 0.1}\left[H_3{O}^{+}\right]=4\times {10}^{-6}pH=-log\left[H_3{O}^{+}\right]pH=-\log (4\times {10}^{-6})=(6-\log (4\left)\right)=(6-2\log (2\left)\right)pH=(6-0.6)=5.4$