Ionic Equilibrium JEE 2018: Dilution Question Solved

What is the ionic equilibrium JEE 2018 dilution question?

The ionic equilibrium JEE 2018 dilution question has option D as its answer: dilution leaves hydrogen-ion concentration unchanged in P and S, increases it in Q and decreases it in R. The source is JEE Advanced 2018, Paper 2, Chemistry, Q.18; “medium difficulty” and the 240-second target are question-bank guidance, not official exam specifications.

Match each preparation with its hydrogen-ion concentration change after adding water and allowing equilibrium to re-establish:

P: Mix 10 mL of 0.1 M NaOH with 20 mL of 0.1 M acetic acid, then make the total volume 60 mL.
Q: Mix 20 mL each of 0.1 M NaOH and 0.1 M acetic acid, then make the total volume 80 mL.
R: Mix 20 mL each of 0.1 M HCl and 0.1 M aqueous ammonia, then make the total volume 80 mL.
S: Dilute 10 mL of saturated nickel(II) hydroxide solution to 20 mL. Excess undissolved nickel(II) hydroxide remains at final equilibrium.

Weak-acid/base dissociation and salt hydrolysis are small. Initial means after mixing and reaction, but before additional dilution. Define the effects by:

f = \frac{[H^{+}]_{n e w}}{[H^{+}]_{o l d}}

1 : f = 1; 2 : f = \frac{1}{2}; 3 : f = 2; 4 : f = \frac{1}{\sqrt{2}}; 5 : f = \sqrt{2}

The options, in P, Q, R, S order, are:

Why does P retain the same hydrogen-ion concentration?

P becomes a buffer with equal amounts of acetic acid and acetate, so dilution leaves their concentration ratio unchanged. Neutralise first, then apply the buffer formula.

C H_{3} C O O H + O H^{-} \to C H_{3} C O O^{-} + H_{2} O

n (N a O H) = 0.1 \times 10 \times 10^{- 3} = 1 \times 10^{- 3} m o l

n (C H_{3} C O O H) = 0.1 \times 20 \times 10^{- 3} = 2 \times 10^{- 3} m o l

After reaction, acetic acid remaining and acetate formed are each 1 mmol:

n (C H_{3} C O O H) = n (C H_{3} C O O^{-}) = 1 \times 10^{- 3} m o l

Neither reagent concentration can be used unchanged. The buffer formula cancels the common solution volume:

[H^{+}] = K_{a} (C H_{3} C O O H) \frac{[C H_{3} C O O H]}{[C H_{3} C O O^{-}]} = K_{a} (C H_{3} C O O H) \frac{10^{- 3} / V}{10^{- 3} / V} = K_{a} (C H_{3} C O O H)

Diluting from 30 mL to 60 mL halves both component concentrations, but their ratio remains one. Within the stated small-dissociation approximation:

\frac{[H^{+}]_{n e w}}{[H^{+}]_{o l d}} = 1, P \to 1

This cancellation does not make buffer invariance exact under unlimited dilution.

Why does Q increase hydrogen-ion concentration on dilution?

Acetate hydrolysis produces hydroxide, whose concentration falls on dilution. Hydrogen-ion concentration therefore rises: convert using the water ion product before matching.

n (N a O H) = n (C H_{3} C O O H) = 0.1 \times 20 \times 10^{- 3} = 2 \times 10^{- 3} m o l

Complete neutralisation leaves 2 mmol of sodium acetate, not an acid–salt buffer. Its analytical salt concentrations are:

C_{i n i t i a l} = \frac{0.002}{0.040} = 0.050 M, C_{f i n a l} = \frac{0.002}{0.080} = 0.025 M

C H_{3} C O O^{-} + H_{2} O ⇌ C H_{3} C O O H + O H^{-}

x = [O H^{-}], K_{b, a c e t a t e} = \frac{x^{2}}{C - x} \approx \frac{x^{2}}{C}

[O H^{-}] = \sqrt{K_{b, a c e t a t e} C}

This is acetate’s base constant, not ammonia’s. Halving salt concentration gives:

\frac{[O H^{-}]_{n e w}}{[O H^{-}]_{o l d}} = \sqrt{\frac{C_{f i n a l}}{C_{i n i t i a l}}} = \frac{1}{\sqrt{2}}

At constant temperature, the water ion product is unchanged. Convert to the requested ion by taking the reciprocal concentration ratio:

[H^{+}] = \frac{K_{w}}{[O H^{-}]}

\frac{[H^{+}]_{n e w}}{[H^{+}]_{o l d}} = \frac{[O H^{-}]_{o l d}}{[O H^{-}]_{n e w}} = \sqrt{2}, Q \to 5

A basic solution moving towards neutrality has increasing hydrogen-ion concentration. Falling salt concentration does not mean every ion concentration falls.

Why does R decrease hydrogen-ion concentration on dilution?

Ammonium hydrolysis produces hydrogen ions directly, and dilution lowers their concentration. Neutralisation forms ammonium chloride, a salt of a weak base and a strong acid:

n (H C l) = n (N H_{3}) = 0.1 \times 20 \times 10^{- 3} = 2 \times 10^{- 3} m o l

H^{+} + N H_{3} \to N H_{4}^{+}

C_{i n i t i a l} = \frac{0.002}{0.040} = 0.050 M, C_{f i n a l} = \frac{0.002}{0.080} = 0.025 M

Ammonium reacts with water to form hydronium. Use the question’s hydrogen-ion notation in the hydrolysis expression:

N H_{4}^{+} + H_{2} O ⇌ N H_{3} + H_{3} O^{+}

x = [H^{+}], K_{a, a m m o n i u m} = \frac{x^{2}}{C - x} \approx \frac{x^{2}}{C}

[H^{+}] = \sqrt{K_{a, a m m o n i u m} C} = \sqrt{\frac{K_{w}}{K_{b, a m m o n i a}} C}

The salt concentration halves, giving the same square-root dependence as Q. Here, however, the formula already gives hydrogen-ion concentration:

\frac{[H^{+}]_{n e w}}{[H^{+}]_{o l d}} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}, R \to 4

Do not take a reciprocal: unlike Q, no change of ion is needed.

Why does S return to its original hydrogen-ion concentration?

Undissolved solid restores the same saturated equilibrium after dilution. Water initially lowers dissolved-ion concentrations, but more solid then dissolves: dissolved solute moles are not fixed.

N i (O H)_{2} (s) ⇌ N i^{2 +} + 2 O H^{-}

K_{s p} = [N i^{2 +}] [O H^{-}]^{2}

At constant temperature, the solubility product is unchanged. Once saturation is restored, hydroxide concentration returns to its original value, and so does hydrogen-ion concentration:

[H^{+}] = \frac{K_{w}}{[O H^{-}]}, f = 1, S \to 1

The answer concerns final equilibrium, not the instant before additional solid dissolves. The four controlling equilibria give:

P, buffer: $f = 1$
Q, basic salt: $f = \sqrt{2}$
R, acidic salt: $f = \frac{1}{\sqrt{2}}$
S, saturated solution with excess solid: $f = 1$

Therefore, the correct option is D:

P \to 1, Q \to 5, R \to 4, S \to 1

How can incorrect ion tracking produce option C?

Option C can result from confusing the hydrolysis ions and ignoring the remaining solid. One possible erroneous derivation is:

Keeping P matched to effect 1 is correct.
Reporting acetate’s hydroxide factor as the hydrogen-ion factor incorrectly matches Q to effect 4.
Taking an unnecessary reciprocal of ammonium’s hydrogen-ion factor incorrectly matches R to effect 5.
Holding dissolved nickel(II) hydroxide moles fixed makes hydroxide halve on doubling volume and hydrogen-ion concentration double, incorrectly matching S to effect 3. This ignores further dissolution.

Those entries give (1, 4, 5, 3), option C. Repair the method: label the ion beside every square root, use the water ion product only when changing ions, and check whether solid can replenish the solution.

Which three practice variations check this method?

Use an unequal buffer, fourfold salt dilution and threefold saturated-solution dilution to test the same method. These are original practice variations, not additional verified PYQs; retain constant temperature, small dissociation/hydrolysis and negligible water autoionisation where relevant.

Unequal buffer amounts: Mix 10 mL of 0.1 M NaOH with 30 mL of 0.1 M acetic acid, then dilute from 40 mL to 80 mL. Find hydrogen-ion concentration in terms of the acid constant and its dilution factor.

Answer: Neutralisation leaves 2 mmol acid and forms 1 mmol acetate. Before and after dilution:

[H^{+}] = 2 K_{a}, f = 1

Fourfold salt dilution: Separately dilute 0.050 M sodium acetate and 0.050 M ammonium chloride to four times their original volumes. Find each final-to-initial hydrogen-ion ratio.

Answer: Acetate needs the reciprocal hydroxide relation; ammonium gives hydrogen ions directly.

f_{a c e t a t e} = \sqrt{4} = 2, f_{a m m o n i u m} = \frac{1}{\sqrt{4}} = \frac{1}{2}

Threefold saturated-solution dilution: Dilute saturated nickel(II) hydroxide solution to three times its volume, with enough excess solid remaining after equilibration. Find the final hydrogen-ion factor.

Answer: The same saturated equilibrium returns, so the factor is one. Remaining solid is the decisive condition.

Before calculating, write three checks: what remains after reaction, which equilibrium controls it, and which ion the formula gives.

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Frequently asked questions

What is the answer to the ionic equilibrium JEE 2018 dilution question?

The correct answer to JEE Advanced 2018, Paper 2, Chemistry, Q.18 is option D: (1, 5, 4, 1). The final-to-initial hydrogen-ion concentration ratios for P, Q, R and S are respectively 1, √2, 1/√2 and 1.

Why does hydrogen-ion concentration stay unchanged in P?

Neutralisation leaves equal amounts of acetic acid and acetate, forming a buffer. Dilution halves both concentrations without changing their ratio, so hydrogen-ion concentration remains equal to the acid dissociation constant within the stated small-dissociation approximation.

Why does diluting sodium acetate increase hydrogen-ion concentration?

Acetate hydrolysis produces hydroxide ions, whose concentration is proportional to the square root of the salt concentration under the small-hydrolysis approximation. Halving the salt concentration reduces hydroxide concentration by a factor of 1/√2. Since [H+] = Kw/[OH−] at constant temperature, hydrogen-ion concentration increases by a factor of √2.

How does dilution affect hydrogen-ion concentration in ammonium chloride?

Ammonium hydrolysis produces hydrogen ions directly, with [H+] proportional to the square root of the salt concentration under the small-hydrolysis approximation. Doubling the solution volume therefore reduces hydrogen-ion concentration by a factor of 1/√2. No reciprocal conversion is needed because the hydrolysis formula already gives [H+].

Why does saturated nickel hydroxide retain the same pH after dilution?

At constant temperature, excess undissolved nickel(II) hydroxide dissolves after water is added and restores the original saturated equilibrium. Provided some solid remains at final equilibrium, hydroxide concentration returns to its original value, so hydrogen-ion concentration and pH also return to their original values.