JEE Advanced 2018 · Paper 2 · Chemistry Q18

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ChemistryList match+3 / -1Question 36 of 54

Dilution processes of different aqueous solutions, with water, are given in LIST-I. The effects of dilution of the solutions on $[\mathrm{H^+}]$ are given in LIST-II.

(Note: Degree of dissociation ( $\alpha$ ) of weak acid and weak base is $\ll 1$ ; degree of hydrolysis of salt $\ll 1$ ; $[\mathrm{H^+}]$ represents the concentration of $\mathrm{H^+}$ ions)

Match each process given in LIST-I with one or more effect(s) in LIST-II. The correct option is

LIST-I

P
(10 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 60 mL
Q
(20 mL of 0.1 M NaOH + 20 mL of 0.1 M acetic acid) diluted to 80 mL
R
(20 mL of 0.1 M HCl + 20 mL of 0.1 M ammonia solution) diluted to 80 mL
S
10 mL saturated solution of $\mathrm{Ni(OH)_2}$ in equilibrium with excess solid $\mathrm{Ni(OH)_2}$ is diluted to 20 mL (solid $\mathrm{Ni(OH)_2}$ is still present after dilution).

LIST-II

1
the value of $[\mathrm{H^+}]$ does not change on dilution
2
the value of $[\mathrm{H^+}]$ changes to half of its initial value on dilution
3
the value of $[\mathrm{H^+}]$ changes to two times of its initial value on dilution
4
the value of $[\mathrm{H^+}]$ changes to $\dfrac{1}{\sqrt{2}}$ times of its initial value on dilution
5
the value of $[\mathrm{H^+}]$ changes to $\sqrt{2}$ times of its initial value on dilution

P → 4; Q → 2; R → 3; S → 1

P → 4; Q → 3; R → 2; S → 3

P → 1; Q → 4; R → 5; S → 3

P → 1; Q → 5; R → 4; S → 1

Answer and worked solution

Official answer: D

P. 1 mmol NaOH with 2 mmol AcOH gives 1 mmol AcOH plus 1 mmol AcO⁻, a buffer with pH = p $K_a$ . Dilution leaves the ratio unchanged, so $[\ce{H+}]$ does not change: 1.
Q. 2 mmol each gives 2 mmol NaOAc. Diluting from 40 to 80 mL halves $C$ . For the salt of a weak acid, $[\ce{H+}]=\sqrt{K_wK_a/C}$ , so $[\ce{H+}]$ rises by $\sqrt2$ : 5.
R. 2 mmol each gives 2 mmol $\ce{NH4Cl}$ , with $C$ halved. For the salt of a weak base, $[\ce{H+}]=\sqrt{K_wC/K_b}$ , so $[\ce{H+}]$ falls to $1/\sqrt2$ of its value: 4.
S. Excess solid $\ce{Ni(OH)2}$ keeps the solution saturated. $[\ce{OH-}]$ , and so $[\ce{H+}]$ , is unchanged: 1.

P → 1; Q → 5; R → 4; S → 1.

(D)

Topics:Hydrolysis of salts Buffer solutions Solubility product

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