At 298 K take Ka(CH3COOH)=1.8×10^-5…

At 298 K take Ka(CH3COOH)=1.8×10^-5 and Kb(NH3)=1.8×10^-5 (so pKa=4.74 for acetic acid). Match the solutions in List-I with their pH in List-II and…

Single correct option

298\ \mathrm{K}

take

K_a(\mathrm{CH_3COOH})=1.8\times10^{-5}

and

K_b(\mathrm{NH_3})=1.8\times10^{-5}

(so

\mathrm{p}K_a=4.74

for acetic acid). Match the solutions in List-I with their

\mathrm{pH}

in List-II and choose the correct option.

List-I
P.

0.10\ \mathrm{M}

sodium acetate
Q.

0.10\ \mathrm{M}

ammonium chloride
R.

100\ \mathrm{mL}

of a buffer that is

0.10\ \mathrm{M}

in both acetic acid and sodium acetate, after adding

10\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{HCl}

0.10\ \mathrm{M}

ammonium acetate

List-II
1.

4.66

5.13

7.00

8.87

2.87

Show answer and solution

Answer: (B) P→4, Q→2, R→1, S→3

P: salt of weak acid and strong base,

\mathrm{pH}=7+\tfrac12(\mathrm{p}K_a+\log c)=7+\tfrac12(4.74-1)=8.87

: P→4. Q:

\mathrm{NH_4^+}

has

\mathrm{p}K_a=14-\mathrm{p}K_b=9.26

, so

\mathrm{pH}=\tfrac12(\mathrm{p}K_a-\log c)=\tfrac12(9.26+1)=5.13

: Q→2. R: the buffer holds

10

mmol of each component; the added

1.0

mmol of

\mathrm{H^+}

converts

1.0

mmol of acetate to acid, giving

11

mmol acid and

9

mmol acetate, so

\mathrm{pH}=4.74+\log(9/11)=4.66

: R→1. S: salt of a weak acid and a weak base,

\mathrm{pH}=7+\tfrac12(\mathrm{p}K_a-\mathrm{p}K_b)=7.00

because the two constants are equal: S→3. The value 2.87 corresponds to

0.10\ \mathrm{M}

acetic acid itself and is unused. Correct code: P→4, Q→2, R→1, S→3.

Original question written in the JEE Advanced 2018 Paper 2 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake