Question

Trimethylamine, $(C{H}_3{)}_{3}N$ is a weak base $K_b=6.4\times {10}^{-5}$ that hydrolyzes by the following equilibrium reaction:
$(C{H}_3{)}_{3}N+H_{2}O\rightleftharpoons (C{H}_3{)}_{3}N{H}^{+}+O{H}^{-}$
The pH of 0.1 M solution of $(C{H}_3{)}_{3}N{H}^{+}$ is:
(Given $log\left(4\right)=0.60$)

• A. 5.4
• B. 8.4
• C. 3.4
• D. 7.4

Answer 1

The correct option is A 5.4

$\left(C{H}_3{)}_{3}N{H}^{+}\right(aq)+H_{2}O(l)\rightleftharpoons (C{H}_3{)}_{3}N\left(aq\right)+H_3{O}^{+}\left(aq\right)Initially:c00Change:-c\alpha c\alpha c\alpha Equilibrium:c-c\alpha c\alpha c\alpha$
We know , aqueous solution at room temperature has,
$K_a\times K_b=K_{w}Here,K_w={10}^{-14}\to IonisationConstantfor{H}_{2}O{K}_a=\frac{K_w}{K_b}{K}_a=\frac{{10}^{-14}}{6.4\times {10}^{-5}}=1.56\times {10}^{-10}$
By definition, the dissociation constant will be :
$K_a=\frac{\left[\right(C{H}_3{)}_{3}N]\times [H_3{O}^{+}]}{\left[\right(C{H}_3{)}_{3}N{H}^{+}]}=\frac{c\alpha \times c\alpha }{c-c\alpha }$
$K_a=\frac{0.1\alpha \times 0.1\alpha }{0.1-0.1\alpha }$
Since dissociation of weak base is less so, $0.1-0.1\alpha \approx 0.1$
$\therefore K_a=\frac{(0.1{)}^2{\alpha }^2}{0.1}=1.56\times {10}^{-10}$$K_a=\left(0.1\right){\alpha }^2=1.56\times {10}^{-10}$${\alpha }^2=1.56\times {10}^{-9}$
$\alpha =\sqrt{1.56\times {10}^{-9}}=3.95\times {10}^{-5}$
$\left[H_3{O}^{+}\right]=c\alpha =0.1\times 3.95\times {10}^{-5}=3.95\times {10}^{-6}$
$pH=-log\left[H_3{O}^{+}\right]\Rightarrow pH=-log(3.95\times {10}^{-6})=6-log3.95\approx 6-log4=6-0.6=5.4$