The ionisation constant of dimethylamine is `5.4xx10^(-4)`. Calculate its degree of ionization in its `0.02M` solution. What percentage of dimethylamine is ionized if the solution is also `0.1 M` in `NaOH`?

Explanation

`alpha=sqrt(K_b/c)`
`K_b=5.4xx10^(-4)`, c=0.02 M
`therefore alpha =sqrt((5.4xx10^(-4))/(2xx10^(-2)))`=0.164
Let the amount of dimethylamine dissociated in the presence of 0.1 M NaOH be x.
`{:(,(CH_3)_2NH+H_2O hArr , (CH_3)_2NH_2^(+)+,OH^(-)),("Initial" , "0.02",,),("After" , 0.02 -x , x, 0.1+x),("dissociation", approx 0.02 , , approx 0.1):}`
`K_b=(x xx(0.1 +x))/(0.02-x) approx (x xx 0.1)/0.02`
or `x=(5.4xx10^(-4)xx0.02)/0.1`
`=1.08xx10^(-4)`
`therefore` % of dimethylamine ionized
`=(1.08xx10^(-4))/0.02xx100` =0.54 %