Question

The ionization constant of dimethylamine is $5.4\times {10}^{-4}$. Calculate its degree of ionization in its 0.02 M solution. What percentage of dimethylamine is ionized if the solution is also 0.1M in $NaOH$?

Answer 1

The degree of ionization, $y=\sqrt{\frac{K_b}{c}}=\sqrt{\frac{5.4\times {10}^{-4}}{2\times {10}^{-2}}}=0.164$
Let x be the amount of dimethyl aniline dissociated in presence of 0.1 M NaOH.
$(C{H}_3{)}_{2}NH+H_{2}O\rightleftharpoons (C{H}_3{)}_{2}N{H}_2^{+}+O{H}^{-}$
The equilibrium concentrations are:
$\left[\right(C{H}_3{)}_{2}NH]=0.02-x\approx 0.02$
$\left[\right(C{H}_3{)}_{2}N{H}_2^{+}]=x$
$\left[O{H}^{-}\right]=0.1+x\approx 0.1$
The equilibrium constant expression is:
$K_b=\frac{\left[\right(C{H}_3{)}_{2}N{H}_2^{+}\left]\right[O{H}^{-}]}{\left[\right(C{H}_3{)}_{2}NH]}=\frac{x\times 0.1}{0.02}$
$x=1.08\times {10}^{-4}$
The percentage of dimethylamine ionized is $\frac{1.08\times {10}^{-4}}{0.02}\times 100=0.54$%.