Question

Percentage ionization of $0.02M$ dimethyl amine in the presence of $1MNaOH$, is $\left(K_b=5.4\times 10^{-4}\right)$

• A. $0.54\%$
• B. $0.17\%$
• C. $0.0054$
• D. $0.0017$

Answer 1

Answer: (A)

${\left(C{H}_3\right)}_{2}NH+H_{2}O\rightleftharpoons {\left(C{H}_3\right)}_{2}N{H}_2^{+}+O{H}^{-}$



$K_b=\frac{\left[{\left(C{H}_3\right)}_{2}N{H}_2^{+}\right]\left[O{H}^{-}\right]}{\left[{\left(C{H}_3\right)}_{2}NH\right]}$

$5.4\times 10^{-4}=\frac{C\alpha \times 0.1}{C(1-\alpha )}=\alpha (0.1)$ (if $\alpha <<1)$

$\therefore \alpha =5.4\times 10^{-3}$

$\%$ of ionization $=5.4\times 10^{-3}\times 100=0.54$