Match the aqueous solutions in List-I…

Match the aqueous solutions in List-I (at 298 K; Ka(CH3COOH)=Kb(NH3)=1.8×10^-5, Kw=10^-14, log 1.8=0.255) with their pH in List-II. List-I (P) 20 mL…

Single correct option

Match the aqueous solutions in List-I (at

298\ \mathrm{K}

;

K_a(\mathrm{CH_3COOH})=K_b(\mathrm{NH_3})=1.8\times10^{-5}

K_w=10^{-14}

\log1.8=0.255

) with their pH in List-II.

List-I
(P)

20\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{CH_3COOH}

mixed with

10\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{NaOH}

(Q)

0.10\ \mathrm{M}

\mathrm{CH_3COONa}

(R)

0.10\ \mathrm{M}

\mathrm{NH_4Cl}

(S)

50\ \mathrm{mL}

0.20\ \mathrm{M}

\mathrm{HCl}

mixed with

50\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{NaOH}

List-II
(1)

5.13

(2)

2.87

(3)

8.87

(4)

4.74

(5)

1.30

Show answer and solution

Answer: (A) P→4, Q→3, R→1, S→5

P: after mixing,

2.0

mmol acid and

1.0

mmol

\mathrm{NaOH}

leave

1.0

mmol

\mathrm{CH_3COOH}

and

1.0

mmol

\mathrm{CH_3COO^-}

, a buffer with

\mathrm{pH}=\mathrm{p}K_a=4.74

(4). Q: salt of a weak acid,

\mathrm{pH}=7+\tfrac12\mathrm{p}K_a+\tfrac12\log C=7+2.37-0.50=8.87

(3). R: salt of a weak base,

\mathrm{pH}=7-\tfrac12\mathrm{p}K_b-\tfrac12\log C=7-2.37+0.50=5.13

(1). S:

10

mmol

\mathrm{HCl}

less

5

mmol

\mathrm{NaOH}

leaves

5

mmol in

100\ \mathrm{mL}

, so

[\mathrm{H^+}]=0.05\ \mathrm{M}

and

\mathrm{pH}=1.30

(5). Value (2),

2.87

, is the pH of

0.10\ \mathrm{M}

acetic acid alone and is not used. The correct code is option A; B swaps Q and R, C confuses P with free acetic acid, D swaps R and S.

Original question written in the JEE Advanced 2025 Paper 1 pattern and checked twice for the correct answer. Not an official question paper. How we write and check questions · Corrections policy · Report a mistake