Equilibrium NEET 2026: Ranking Salts by Molar Solubility

Equilibrium NEET 2026: which salt is most soluble?

Zn(OH)₂ is the most soluble salt in this Equilibrium NEET 2026 question, even though AgBr has the largest solubility product. The salts release different numbers of ions, so ranking their solubility products directly gives the wrong answer.

Source: NEET 2026, held on 21 June 2026, Code-50, Chemistry, Equilibrium. Medium difficulty is the question bank’s classification, not an official exam rating.

Arrange these salts in decreasing molar solubility in water at 298 K, using the supplied constants:

AgBr:

K_{s p} = 5.0 \times 10^{- 13}

Zn(OH)₂:

K_{s p} = 1.0 \times 10^{- 15}

Hg₂Cl₂:

K_{s p} = 1.3 \times 10^{- 18}

Choose from:

H g_{2} C l_{2} > Z n (O H)_{2} > A g B r

A g B r > Z n (O H)_{2} > H g_{2} C l_{2}

H g_{2} C l_{2} > A g B r > Z n (O H)_{2}

Z n (O H)_{2} > A g B r > H g_{2} C l_{2}

Answer: D. The worked solution below explains why.

How do you write the ion concentrations before choosing a formula?

Start with the balanced dissolution equation, then record how much of each ion one mole of salt produces. Molar solubility measures dissolved salt formula units, not the concentration of every individual ion. Define it as:

s = molar solubility of the salt, in m o l L^{- 1}

AgBr: one mole gives one mole of each ion.

A g B r (s) ⇌ A g^{+} (a q) + B r^{-} (a q)

[A g^{+}] = s, [B r^{-}] = s

K_{s p} = [A g^{+}] [B r^{-}] = s \times s = s^{2}

Zn(OH)₂: one mole gives twice as much hydroxide as zinc ion.

Z n (O H)_{2} (s) ⇌ Z n^{2 +} (a q) + 2 O H^{-} (a q)

[Z n^{2 +}] = s, [O H^{-}] = 2 s

K_{s p} = [Z n^{2 +}] [O H^{-}]^{2} = s (2 s)^{2} = 4 s^{3}

Hg₂Cl₂: one mole gives one dimeric mercury cation and two chloride ions.

H g_{2} C l_{2} (s) ⇌ H g_{2}^{2 +} (a q) + 2 C l^{-} (a q)

[H g_{2}^{2 +}] = s, [C l^{-}] = 2 s

K_{s p} = [H g_{2}^{2 +}] [C l^{-}]^{2} = s (2 s)^{2} = 4 s^{3}

The mercury ion is one dimeric cation. Its internal subscript does not mean dissolution releases two independent mercury cations. Use this ion-concentration ledger rather than choosing a formula from the salt’s appearance.

How do you calculate all three molar solubilities?

Take a square root for AgBr and cube roots for the other two salts. Before evaluating each root, rewrite the power of ten as a perfect square or cube. This follows the supplied official solution and keeps the arithmetic checkable without a calculator.

AgBr: substitute its constant into the squared relationship. $s^{2} = 5.0 \times 10^{- 13}$

s = \sqrt{5.0 \times 10^{- 13}} = \sqrt{50 \times 10^{- 14}} = \sqrt{50} \times 10^{- 7}

s_{A g B r} = 7.07 \times 10^{- 7} m o l L^{- 1}

The exponent is now even, so its square root is immediate. Keep the coefficient to two decimal places because the mercury result is close.

Zn(OH)₂: divide the solubility product by four before taking the cube root. $4 s^{3} = 1.0 \times 10^{- 15}$

s^{3} = \frac{1.0 \times 10^{- 15}}{4} = 2.5 \times 10^{- 16} = 250 \times 10^{- 18}

s = \sqrt[3]{250} \times 10^{- 6} \approx 6.30 \times 10^{- 6} m o l L^{- 1}

Check the cube-root coefficient: ${6.30}^{3} = 250.047 \approx 250$

The division by four comes from squaring the hydroxide concentration, including its coefficient. Omitting that factor changes the calculated solubility.

Hg₂Cl₂: use the same cubic relationship with its own constant. $4 s^{3} = 1.3 \times 10^{- 18}$

s^{3} = \frac{1.3 \times 10^{- 18}}{4} = 3.25 \times 10^{- 19} = 325 \times 10^{- 21}

s = \sqrt[3]{325} \times 10^{- 7} \approx 6.87 \times 10^{- 7} m o l L^{- 1}

Check the supplied official approximation: ${6.87}^{3} = 324.242703 \approx 324$

That is close to 325, so the quoted approximation is sufficient for this comparison. More decimal places are unnecessary to separate mercury(I) chloride from AgBr.

How do you compare AgBr and Hg₂Cl₂ without rounding away the difference?

Put all three answers on the same power-of-ten scale, then compare their coefficients. Zn(OH)₂ is first; AgBr is slightly more soluble than Hg₂Cl₂.

Zn(OH)₂:

s_{Z n (O H)_{2}} = 63.0 \times 10^{- 7} m o l L^{- 1}

AgBr:

s_{A g B r} = 7.07 \times 10^{- 7} m o l L^{- 1}

Hg₂Cl₂:

s_{H g_{2} C l_{2}} = 6.87 \times 10^{- 7} m o l L^{- 1}

The coefficients give: $63.0 > 7.07 > 6.87$

Therefore:

Z n (O H)_{2} > A g B r > H g_{2} C l_{2}

Option D is correct. Rounding both smaller solubilities to the value below hides their ordering:

7 \times 10^{- 7} m o l L^{- 1}

Retain enough digits until the final comparison. Here, the two decimal places distinguish the smaller solubilities.

Why does comparing Ksp directly give option B?

Direct comparison gives B because it treats ion-concentration products as molar solubilities. The supplied constants follow:

5.0 \times 10^{- 13} > 1.0 \times 10^{- 15} > 1.3 \times 10^{- 18}

Incorrectly transferring that order to solubility gives:

A g B r > Z n (O H)_{2} > H g_{2} C l_{2} (option B)

The mistake is ignoring the different relationships:

K_{s p} = s^{2} for A g B r

K_{s p} = 4 s^{3} for Z n (O H)_{2} and H g_{2} C l_{2}

Applying the following rule to every salt also preserves the wrong ranking, because square roots preserve the order of positive numbers: $s = \sqrt{K_{s p}}$

Direct solubility-product ranking works for salts with the same dissolution stoichiometry under comparable conditions. Here, it can compare Zn(OH)₂ with Hg₂Cl₂, but it cannot rank all three salts. Check the dissolution equations before using that shortcut.

Which three related questions check whether you understand the method?

Use these original related practice questions, not additional NEET PYQs, to test ion concentration, a valid ranking shortcut and precipitation. Attempt each before reading its worked answer.

Question 1: What is the chloride concentration in saturated Hg₂Cl₂ solution under the same model?

The chloride concentration is twice the salt’s molar solubility. Each dissolved formula unit supplies two chloride ions:

[C l^{-}] = 2 s = 2 \times 6.87 \times 10^{- 7} = 1.374 \times 10^{- 6} m o l L^{- 1}

Question 2: Can Zn(OH)₂ and Hg₂Cl₂ be ranked without taking cube roots?

Yes, Zn(OH)₂ is more soluble because both salts obey the same relationship. Dividing their positive constants by four and taking cube roots preserves their order:

s = {(\frac{K_{s p}}{4})}^{1 / 3}

1.0 \times 10^{- 15} > 1.3 \times 10^{- 18}

Question 3: Will Zn(OH)₂ precipitate when both ion concentrations equal the AgBr solubility calculated above?

No precipitation is expected under this model because the ion product is below the solubility product. Use the independently specified concentrations directly:

[Z n^{2 +}] = [O H^{-}] = 7.07 \times 10^{- 7} m o l L^{- 1}

Q_{s p} = [Z n^{2 +}] [O H^{-}]^{2} = (7.07 \times 10^{- 7})^{3} \approx 3.53 \times 10^{- 19}

Q_{s p} < K_{s p} = 1.0 \times 10^{- 15}

Do not replace the supplied hydroxide concentration with twice the zinc concentration: the dissolution ledger does not override independently specified concentrations. First decide whether ion concentrations must be derived from dissolution or are already given.

Next step: the past-paper archive on NEET JEEnius AI and search past NEET papers by year, subject or chapter, each with a worked solution (100 free searches a month).

Keep going with Simple Pendulum NEET 2026: Finding the Effective Length.

Frequently asked questions

Which is most soluble in water: AgBr, Zn(OH)₂ or Hg₂Cl₂?

Using the supplied Ksp values at 298 K, Zn(OH)₂ has the highest molar solubility. The decreasing order is Zn(OH)₂ > AgBr > Hg₂Cl₂, so option D is correct.

Can I compare Ksp values directly to rank solubility?

Direct Ksp ranking works for salts with the same dissolution stoichiometry under comparable conditions. AgBr obeys Ksp = s², while Zn(OH)₂ and Hg₂Cl₂ obey Ksp = 4s³. You can therefore compare the latter two directly, but not all three.

Why is Ksp equal to 4s³ for Zn(OH)₂?

If the molar solubility is s, dissolution gives a zinc ion concentration of s and a hydroxide ion concentration of 2s. Substituting into the solubility-product expression gives Ksp = s × (2s)² = 4s³. The factor of four comes from squaring the hydroxide concentration.

Does Hg₂Cl₂ release two mercury ions when it dissolves?

No. Each dissolved formula unit releases one dimeric mercury(I) cation, Hg₂²⁺, and two chloride ions. For molar solubility s, their concentrations are s and 2s respectively, giving Ksp = 4s³.