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Equilibrium JEE 2018: Fixed-pH Solubility Solution

JEE Advanced 2018 Chemistry Equilibrium Solubility of salt of weak acid at fixed pH

By Founder, JEEnius - IIT Kanpur Alumni · Nov 2, 2026 · 4 min read

Hard 3 min target

The solubility of a salt of weak acid AB at pH 3 is Y×10−3 mol L−1. The value of Y is _____. Given that the value of solubility product of AB is Ksp=2×10−10 and the value of ionization constant of HB is Ka=1×10−8.

Show answerAnswer

4.47

Explanation

Let the total solubility of AB be S mol L−1. The salt dissociates as:

AB⇌A++B−

Because B− is the conjugate base of weak acid HB, it reacts with hydrogen ions present at pH 3:

B−+H+⇌HB

At pH 3:

[H+]=10−3

Let x be the concentration of B− converted into HB. Then:

[A+]=S

[B−]=S−x

[HB]=x

Using solubility product:

Ksp=[A+][B−]

2×10−10=S(S−x)

For the weak acid HB:

Ka=[H+][B−][HB]

1×10−8=10−3(S−x)x

Rearranging:

xS−x=105

Now multiply this relation with the solubility product relation in the useful form:

S(S−x)=2×10−10

Since:

xS−x=105

We get:

Sx=2×10−5

From the solubility product equation:

S(S−x)=2×10−10

S2−Sx=2×10−10

Substitute Sx=2×10−5:

S2−2×10−5=2×10−10

S2=2×10−5+2×10−10

Since 2×10−10 is very small compared to 2×10−5:

S2≈2×10−5

S=2×10−5

S=4.47×10−3 mol L−1

Given:

S=Y×10−3 mol L−1

Therefore:

Y=4.47

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What is the answer to the equilibrium JEE 2018 fixed-pH solubility question?

The equilibrium JEE 2018 numerical answer is 4.47: total dissolved salt is not the same as free anion concentration. The source is JEE Advanced 2018, Paper 1, Chemistry, Equilibrium. This is a numerical-answer question, not an MCQ.

Paraphrased: a salt of a weak acid dissolves at maintained pH 3. Find the coefficient in its stated molar solubility using these constants:

Salt: AB,S=Y×10−3 molL−1
Ksp=2×10−10,Ka(HB)=1×10−8

The question bank labels this medium difficulty and gives 180 seconds as the expected solving time. That is the question bank's practice estimate, not an official exam time limit.

How do you separate total solubility from free anion concentration?

Total solubility counts all dissolved salt, including the anion converted into weak acid. The positive ion remains free in the stated model, but the anion exists in two forms. Track those forms separately rather than assuming equal free-ion concentrations.

Define the quantities used in the official solution: S=total molar solubility of AB

x=concentration of B− converted into HB

The coupled equilibria are: AB(s)⇌A++B− B−+H+⇌HB

At pH 3:

[H+]=10−3 molL−1

Each mole of dissolved salt releases one mole of each ion. Protonation redistributes the B-containing species, giving:

[A+]=S,[B−]=S−x,[HB]=x

The B-species balance is:

[B−]+[HB]=(S−x)+x=S

Maintained pH means fixed hydrogen-ion concentration. Do not treat it as an initial supply and subtract the amount consumed by protonation. The stated value is its equilibrium concentration.

How do you combine the solubility product and acid constant?

Use the solubility product for the free ions and the acid constant to connect free anion with weak acid. Then multiply the resulting equations to cancel the free-anion term. This eliminates one concentration without a memorised solubility formula.

The solubility product gives:

Ksp=[A+][B−]
S(S−x)=2×10−10

For the weak acid:

Ka=[H+][B−][HB]
1×10−8=10−3(S−x)x

Divide by the fixed hydrogen-ion concentration, then invert:

S−xx=10−810−3=10−5
xS−x=105

The weak-acid concentration is 100,000 times the free-anion concentration. Most dissolved B-containing species are protonated, so free anion cannot represent total solubility.

Multiply the two relations:

S(S−x)×xS−x=(2×10−10)×105

The free-anion concentration is nonzero, so its factor cancels:

S\cancel(S−x)×x\cancelS−x=Sx=2×10−5

How do you calculate the solubility and obtain 4.47?

The required entry is the coefficient 4.47, not the full molar solubility. The official approximation is justified because the discarded contribution to squared solubility is only 0.001% of the dominant term.

Expand the solubility-product equation:

S(S−x)=2×10−10

S2−Sx=2×10−10 Substitute the product obtained above:

S2−2×10−5=2×10−10
S2=2×10−5+2×10−10

The relative size of the smaller term is:

2×10−102×10−5=10−5=0.001%

Following the official approximation: S2≈2×10−5

S≈2×10−5≈4.47×10−3 molL−1=0.00447 molL−1

Compare this with the requested form:

S=Y×10−3 molL−1
Y=4.47

Keeping the small correction does not change the reported answer at two decimal places. Enter the coefficient, not the full molar solubility.

Why does taking the square root of the solubility product fail?

That method incorrectly treats both free-ion concentrations as equal to total solubility. Equal amounts are released by 1:1 dissolution, but protonation removes free anion and stores it as weak acid. There are no answer options here: the following is an incorrect entry, not an official distractor.

The faulty assumption is: [A+]=[B−]=S

It produces:

S=Ksp=2×10−10≈1.41×10−5 molL−1

Converting this incorrect solubility into the requested format gives:

Y=1.41×10−510−3≈0.0141

The valid balance and invalid equality are:

Valid:[A+]=[B−]+[HB]
Invalid here:[A+]=[B−]

Use a direction check: protonation reduces free anion, so more salt must dissolve to satisfy the solubility product. The calculated solubility must therefore exceed the square-root result. This checks the direction of the change, but does not replace the two-equilibrium derivation.

What three follow-up questions test the same setup?

Find the free-anion fraction, calculate its concentration and predict the effect of increasing pH. These are original follow-up exercises based on the supplied question, not additional JEE PYQs. Each has a worked answer immediately below it.

What fraction of dissolved B-containing species remains as free anion?

Only about one part in 100,000 remains free. Use the ratio already established, then apply the B-species balance: x=105(S−x)

S=(S−x)+x=100001(S−x)
[B−]S=S−xS=1100001≈10−5

What is the equilibrium concentration of free anion?

Divide the solubility product by the positive-ion concentration, which equals total solubility. Using the calculated value gives:

[B−]=Ksp[A+]=2×10−104.47×10−3≈4.47×10−8 molL−1

Does solubility rise or fall when maintained pH increases?

It falls within the stated equilibrium model, with both equilibrium constants unchanged. Increasing pH lowers hydrogen-ion concentration, so less anion is protonated. Less salt then needs to dissolve to satisfy the solubility product.

For your next attempt, cover the solution and write the B-species balance first. Then reproduce the multiplication that eliminates the free-anion term before doing any square-root calculation.

Next step: photograph a doubt on JEEnius and photograph any question you are stuck on and get a step-by-step solution, with a free-body diagram when the question needs one (20 free).

Keep going with Electrostatics JEE 2019: Spherical Shell With a Small Hole.

Frequently asked questions

What is the answer to the JEE Advanced 2018 fixed-pH solubility question?

The required numerical entry is 4.47. The calculated molar solubility is approximately 4.47 × 10⁻³ mol L⁻¹, but the question asks for Y in S = Y × 10⁻³ mol L⁻¹.

Why can't I use S = √Ksp in this question?

That expression assumes both free-ion concentrations equal the total molar solubility. Here, protonation converts most B⁻ into HB, so [A⁺] = S but [B⁻] is much smaller. Use Ksp = [A⁺][B⁻] together with the acid equilibrium and the balance [B⁻] + [HB] = S.

What does maintained pH 3 mean in a solubility problem?

It means the equilibrium hydrogen-ion concentration stays fixed at 10⁻³ mol L⁻¹. Do not subtract the hydrogen ions consumed by protonation from this value.

Does solubility increase or decrease when pH increases?

Solubility decreases in this model when the equilibrium constants remain unchanged. Increasing pH lowers the hydrogen-ion concentration, so less B⁻ is converted into HB. Less salt then needs to dissolve to satisfy the solubility product.

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