Question

Q). In a titration of 1 N weak acid HX with 2 N NaOH. The pH is 5.8 after the addition of 10 ml of NaOH and is 6.8 after the addition of 20 ml of NaOH solution. The ionisation constant of acid is.

Answer 1

Dear Student,


$$\begin{gathered} \mathrm{Let}\mathrm{V}\mathrm{be}\mathrm{the}\mathrm{volume}\mathrm{of}\mathrm{acid}. \\ \mathrm{We}\mathrm{know}\mathrm{that}, \\ \mathrm{pH}={\mathrm{pK}}_{\mathrm{a}}+\hspace{0.17em}\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]} \\ \mathrm{After}\mathrm{addition}\mathrm{of}10\mathrm{ml}\mathrm{NaOH}, \\ \left[\mathrm{Salt}\right]=10\times 1=10 \\ \left[\mathrm{Acid}\right]=\mathrm{V}\times 1-10\times 1=\mathrm{V}-10 \\ \mathrm{So},5.8={\mathrm{pK}}_{\mathrm{a}}+\hspace{0.17em}\log \frac{\left(10\right)}{(\mathrm{V}-10)}...\left(\mathrm{i}\right) \\ \mathrm{After}\mathrm{addition}\mathrm{of}20\mathrm{ml}\mathrm{of}\mathrm{NaOH} \\ \left[\mathrm{Salt}\right]=20\times 1=20 \\ \left[\mathrm{Acid}\right]=\mathrm{V}\times 1-20\times 1=\mathrm{V}-20 \\ \mathrm{So},6.8={\mathrm{pK}}_{\mathrm{a}}+\hspace{0.17em}\log \frac{\left(20\right)}{(\mathrm{V}-20)}..\left(\mathrm{ii}\right) \\ \mathrm{Substracting}\left(\mathrm{i}\right)\mathrm{from}\left(\mathrm{ii}\right) \\ 1=\log [\frac{\left(20\right)}{(\mathrm{V}-20)}\frac{(\mathrm{V}-10)}{\left(10\right)}] \\ \mathrm{or},1=\log \left[\frac{2(\mathrm{V}-10)}{(\mathrm{V}-20)}\right] \\ \frac{2(\mathrm{V}-10)}{(\mathrm{V}-20)}=10 \\ \mathrm{or},\frac{(\mathrm{V}-10)}{(\mathrm{V}-20)}=5 \\ \mathrm{or},\mathrm{V}-10=5\mathrm{V}-100 \\ \mathrm{or},4\mathrm{V}=90 \\ \mathrm{or},\mathrm{V}=\frac{90}{4}=22.5\mathrm{ml} \\ \mathrm{Substituting}\mathrm{in}\left(\mathrm{i}\right) \\ {\mathrm{pK}}_{\mathrm{a}}=5.8-\log \left(\frac{\left(10\right)}{(22.5-10)}\right) \\ =5.8+0.096 \\ =5.89 \end{gathered}$$