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QJEEAdv18Ch7d7e: JEE Advanced 2018 Chemistry - Class 11 JEEAdv18Ch7d7e Solution: Dilution processes of different aqueous solutions, with water,…

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

(Note: Degree of dissociation (a) of weak acid and weak base is <<1; degree of hydrolysis of salt <<1; [H+]\left[ {{H^ + }} \right] represents the concentration of H+{H^ + } ions)

LIST-I LIST-II
P. (10 mL of 0.1 M NaOH + 20 mL of
0.1 M acetic acid) diluted to 60 mL		 1. the value of [H+] does not change
on dilution
Q. (20 mL of 0.1 M NaOH + 20 mL of
0.1 M acetic acid) diluted to 80 mL		 2. the value of [H+] changes to half
of its initial value on dilution
R. (20 mL of 0.1 M HCL + 20 mL of
0.1 M ammonia solution) diluted to
80 mL 		3. the value of [H+] changes to two
times of its initial value on dilution
S. 10 mL saturated solution of Ni(OH)2
in equilibrium with excess solid
Ni(OH)2 is diluted to 20 mL (solid
Ni(OH)2 is still present after dilution).		 4. the value of [H+] changes to 12{1 \over {\sqrt 2 }}
times of its initial value on dilution
5. the value of [H+] changes to 2\sqrt 2
times of its initial value on dilution


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

Multiple Choice Options

  • A P−4;Q−2;R−3;S−1P - 4;Q - 2;R - 3;S - 1
  • B P−4;Q−3;R−2;S−3P - 4;Q - 3;R - 2;S - 3
  • C P−1;Q−4;R−5;S−3P - 1;Q - 4;R - 5;S - 3
  • D P−1;Q−5;R−4;S−1P - 1;Q - 5;R - 4;S - 1

Step-by-Step Verified Solution

P. Sodium hydroxide reacts with acetic acid to form sodium acetate.

JEE Advanced 2018 Paper 2 Offline Chemistry - Ionic Equilibrium Question 9 English Explanation 1

The final solution will be a buffer of CH3COOH\mathrm{CH}_3 \mathrm{COOH} and CH3COONa\mathrm{CH}_3 \mathrm{COONa} as CH3COOH\mathrm{CH}_3 \mathrm{COOH} is not completely neutralised by NaOH\mathrm{NaOH} .

The final moles of CH3COOH=\mathrm{CH}_3 \mathrm{COOH}= moles of CH3COONa\mathrm{CH}_3 \mathrm{COONa} = 0.001 mol

Ph=p Ka+log⁡[CH3COO−][CH3COOH]\mathrm{Ph}=p \mathrm{~K}_a+\log \frac{\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]}{\left[\mathrm{CH}_3 \mathrm{COOH}\right]}

⇒[H+]\Rightarrow\left[\mathrm{H}^{+}\right] will not change on dilution as concentration of salt and acid does not change on dilution.

∴\therefore Correct match P−1\mathrm{P}-1

Q. JEE Advanced 2018 Paper 2 Offline Chemistry - Ionic Equilibrium Question 9 English Explanation 2

amp;[CH3COO−]old=20×0.140=240amp;[CH3COO−]new =280\begin{aligned}& \left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]_{\mathrm{old}}=\frac{20 \times 0.1}{40}=\frac{2}{40} \\\\& {\left[\mathrm{CH}_3 \mathrm{COO}^{-}\right]_{\text {new }}=\frac{2}{80}}\end{aligned}

[OH−]amp;=KHCamp;=(Kw KaC)[H+]1amp;=Kw KaC[H+]2[H+]1=C1C2amp;=0.050.025=2\begin{aligned}{\left[\mathrm{OH}^{-}\right] } & =\sqrt{\mathrm{K}_{\mathrm{H}} \mathrm{C}} \\\\& =\sqrt{\left(\frac{\mathrm{K}_w}{\mathrm{~K}_a} \mathrm{C}\right)} \\\\{\left[\mathrm{H}^{+}\right]_1 } & =\sqrt{\frac{\mathrm{K}_w \mathrm{~K}_a}{\mathrm{C}}} \\\\\frac{\left[\mathrm{H}^{+}\right]_2}{\left[\mathrm{H}^{+}\right]_1}=\sqrt{\frac{\mathrm{C}_1}{\mathrm{C}_2}} & =\sqrt{\frac{0.05}{0.025}}=\sqrt{2}\end{aligned}

Correct match Q-5

R. JEE Advanced 2018 Paper 2 Offline Chemistry - Ionic Equilibrium Question 9 English Explanation 3

As per the condition given in RR the resultant solution before dilution contains 2 millimoles of NH4Cl\mathrm{NH}_4 \mathrm{Cl} in 40 mL40 \mathrm{~mL} of solution. Hence, a salt of strong acid and weak base is formed. For this,

[H+]initial =Kw×CKb\left[\mathrm{H}^{+}\right]_{\text {initial }}=\sqrt{\frac{K_w \times C}{K_b}}

Now on dilution upto 80 mL80 \mathrm{~mL} new conc. becomes C/2C / 2 .

So [H+]new =Kw×C2Kb\left[\mathrm{H}^{+}\right]_{\text {new }}=\sqrt{\frac{K_w \times \frac{C}{2}}{K_b}}

or [H+]new =[H+]initial ×12\left[\mathrm{H}^{+}\right]_{\text {new }}=\left[\mathrm{H}^{+}\right]_{\text {initial }} \times \frac{1}{\sqrt{2}}

(S) For a saturated solution,

JEE Advanced 2018 Paper 2 Offline Chemistry - Ionic Equilibrium Question 9 English Explanation 4
Ksp=s×(2s)2=4s3⇒s=[OH−]=Ksp43\begin{aligned}K_{s p}=s \times(2 s)^2=4 s^3 \\\\\Rightarrow s=\left[\mathrm{OH}^{-}\right]=\sqrt[3]{\frac{K_{s p}}{4}}\end{aligned}

Irrespective of volume of solution, [H+]\left[\mathrm{H}^{+}\right] remains constant.

(S) →\rightarrow (1)

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