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 so

Q: What is the correct answer for JEE Main Chemistry Q56?

The correct answer is 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: $\mathrm{HX} + \mathrm{NaOH} \rightarrow \mathrm{NaA} + \mathrm{H_2O}$ Initial moles: $20 \times 0.2 = 4\mathrm{mmol}$ of HX, $

Q: 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…

The correct answer is 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: $\mathrm{HX} + \mathrm{NaOH} \rightarrow \mathrm{NaA} + \mathrm{H_2O}$ Initial moles: $20 \times 0.2 = 4\mathrm{mmol}$ of HX, $

Q: What are the options for this JEE Main Chemistry question?

A. 7, 2.0 · B. 0, 3.3 · C. 1, 2.2 · D. 0, 2.2

Question

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 : , ,

FirstInTest · Accepted Answer

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. $\mathrm{pH} = \mathrm{pK_a} + \log\frac{[\mathrm{salt}]}{[\mathrm{acid}]}$ $\mathrm{pH} = 3.3 + \log\frac{2\mathrm{mmol}}{2\mathrm{mmol}}$ $\mathrm{pH} = 3.3 + \log(1) = 3.3 + 0 = 3.3$

Answer

7, 2.0

Answer

0, 3.3

Answer

1, 2.2

Answer

0, 2.2

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