Q.56. At , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH i

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Q.56. At , 20.0 mL of 0.2 M weak monoprotic acid HX is titrated against 0.2 M NaOH. The pH of the solution (a) at the start of the titration (when NaOH has not been added) and (b) when 10 mL of NaOH is added respectively, are : Given : , ,

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Correct answer: B. Sol. For 0.2M, 20ml HX solution, (a) At the start of titration (only HX): $\mathrm{pH\;of\;weak\;acid} = \frac{1}{2}[\mathrm{pK_a} - \log\mathrm{c}]$ $= \frac{1}{2}[3.3 - \log(0.2)]$ $= \frac{1}{2}[3.3 - (-0.7)] = \frac{1}{2}[4] = 2$ (b) When 10 mL of 0.2 M NaOH is added: $\mathrm{HX} + \mathrm{NaOH} ightarrow \mathrm{NaA} + \mathrm{H_2O}$ Initial moles: $20 	imes 0.2 = 4\mathrm{mmol}$ of HX, $10 	imes 0.2 = 2\mathrm{mmol}$ of NaOH After reaction: $2\mathrm{mmol}$ of HX, $0\mathrm{mmol}$ of NaOH, $2\mathrm{mmol}$ of NaA This forms an acidic buffer solution. $\mathrm{pH} = \mathrm{pK_a} + \log\frac{[\mathrm{salt}]}{[\mathrm{acid}]}$ $\mathrm{pH} = 3.3 + \log\frac{2\mathrm{mmol}}{2\mathrm{mmol}}$ $\mathrm{pH} = 3.3 + \log(1) = 3.3 + 0 = 3.3$

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