Take pK_a(CH3COOH) = 4.74 and pK_b(NH3) = 4.74. Which of the following mixtures give a solution with the pH stated?

100 mL of 0.10 M HCl + 100 mL of 0.20 M CH3COONa; pH = 4.74

100 mL of 0.10 M CH3COOH + 50 mL of 0.10 M NaOH; pH = 4.74

100 mL of 0.20 M NH3 + 50 mL of 0.20 M HCl; pH = 9.26

100 mL of 0.10 M HCl + 100 mL of 0.20 M CH3COONa; pH = 4.74

100 mL of 0.10 M CH3COOH + 100 mL of 0.10 M NaOH; pH = 4.74

100 mL of 0.10 M CH3COOH + 50 mL of 0.10 M NaOH; pH = 4.74

100 mL of 0.20 M NH3 + 50 mL of 0.20 M HCl; pH = 9.26

JEE Advanced levelOne or more correct optionsDifficulty 3 of 3ia219

Take $p K_{a} (CH X_{3} COOH) = 4.74$ and $p K_{b} (NH X_{3}) = 4.74$ . Which of the following mixtures give a solution with the pH stated?

A $100 mL$ of $0.10 M$ $HCl$ + $100 mL$ of $0.20 M$ $CH X_{3} COONa$ ; pH $= 4.74$
B $100 mL$ of $0.10 M$ $CH X_{3} COOH$ + $100 mL$ of $0.10 M$ $NaOH$ ; pH $= 4.74$
C $100 mL$ of $0.10 M$ $CH X_{3} COOH$ + $50 mL$ of $0.10 M$ $NaOH$ ; pH $= 4.74$
D $100 mL$ of $0.20 M$ $NH X_{3}$ + $50 mL$ of $0.20 M$ $HCl$ ; pH $= 9.26$

(a)

10

mmol acid,

5

mmol base leave

5

mmol each of

CH X_{3} COOH

and

CH X_{3} COO X^{-}

, so pH

= p K_{a} = 4.74

. (b) Equal moles give pure

CH X_{3} COONa

, which is basic (pH about

8.7

), not

4.74

. (c)

20

mmol

NH X_{3}

and

10

mmol

HCl

leave

10

mmol each of

NH X_{3}

and

NH X_{4} X^{+}

pOH = 4.74

, pH

= 9.26

. (d)

10

mmol

HCl

converts

10

20

mmol acetate into acid, leaving

10

mmol each, so pH

= 4.74