Aqueous solutions are prepared at 298…

Aqueous solutions are prepared at 298 K using a weak acid HA with Ka=1.0×10^-5 and a weak base BOH with Kb=1.0×10^-5. Match the solution in List-I…

Single correct option

Aqueous solutions are prepared at

298\ \mathrm{K}

using a weak acid

\mathrm{HA}

with

K_a=1.0\times10^{-5}

and a weak base

\mathrm{BOH}

with

K_b=1.0\times10^{-5}

. Match the solution in List-I with its pH (nearest integer) in List-II. (

K_w=1.0\times10^{-14}

)

List-I: (P)

0.10\ \mathrm{M}

\mathrm{HA}

; (Q)

0.10\ \mathrm{M}

\mathrm{NaA}

; (R)

100\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{HA}

mixed with

50\ \mathrm{mL}

0.10\ \mathrm{M}

\mathrm{NaOH}

; (S)

0.10\ \mathrm{M}

\mathrm{BA}

(salt of

\mathrm{BOH}

and

\mathrm{HA}

)

List-II: (1)

3

; (2)

5

; (3)

7

; (4)

9

; (5)

11

Show answer and solution

Answer: (D) P→1, Q→4, R→2, S→3

(P) Weak acid:

[\mathrm{H^+}]=\sqrt{K_aC}=\sqrt{10^{-5}\times0.1}=10^{-3}

, pH

=3

\Rightarrow

1.
(Q) Salt of weak acid and strong base:

K_h=K_w/K_a=10^{-9}

[\mathrm{OH^-}]=\sqrt{10^{-9}\times0.1}=10^{-5}

, pOH

=5

, pH

=9

\Rightarrow

4.
(R)

10\ \mathrm{mmol}

\mathrm{HA}

and

5\ \mathrm{mmol}

\mathrm{NaOH}

give

5\ \mathrm{mmol}

\mathrm{HA}

and

5\ \mathrm{mmol}

\mathrm{A^-}

(buffer): pH

=\mathrm{p}K_a+\log\dfrac{[\mathrm{A^-}]}{[\mathrm{HA}]}=5+0=5

\Rightarrow

2.
(S) For a salt of a weak acid and a weak base, pH

=7+\tfrac12(\mathrm{p}K_a-\mathrm{p}K_b)=7+0=7

since

K_a=K_b

\Rightarrow

3.
(Exact quadratic solutions give pH

3.002

for P and

8.99998

for Q.) Correct code: P→1, Q→4, R→2, S→3.

Original question written in the JEE Advanced 2023 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