Ionic Equilibrium JEE 2025: Barium Iodate Solubility

What is the answer to the Ionic Equilibrium JEE 2025 barium iodate question?

3.95 is the required numerical entry for this Ionic Equilibrium JEE 2025 question. Use precipitation first, equilibrium second: calculate the iodate left after reaction before applying the solubility product.

This is from JEE Advanced 2025, Paper 2, Chemistry, in numerical-answer format. Mix 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate, then find the molar solubility of barium iodate in the resulting mixture. Use the given solubility product and report the requested coefficient:

K_{s p} = 1.58 \times 10^{- 9}, s = X \times 10^{- 6} m o l d m^{- 3}

The question bank tags it medium and gives an expected solving time of 180 seconds. Neither is an official exam classification.

How do you count the ions and identify what precipitates?

Barium iodate precipitates, with barium ions limiting the reaction and iodate ions remaining in excess. Start with moles, not concentrations: the balanced reaction fixes how much of each ion is consumed.

B a (N O_{3})_{2} (a q) + 2 N a I O_{3} (a q) \to B a (I O_{3})_{2} (s) + 2 N a N O_{3} (a q)

Convert the volumes before multiplying by the stock concentrations. Each salt supplies one mole of its reacting ion per mole of salt:

200 m L = 0.200 L, 100 m L = 0.100 L

n_{B a^{2 +}} = 0.010 \times 0.200 = 0.002 m o l = 2 m m o l

n_{I O_{3}^{-}} = 0.10 \times 0.100 = 0.010 m o l = 10 m m o l

One mole of barium ions requires two moles of iodate ions. Therefore, 2 mmol of barium ions consumes 4 mmol of iodate, leaving iodate in excess.

Treat barium as consumed for this stoichiometric bookkeeping. Its final equilibrium concentration is not literally zero: the equilibrium calculation determines the small dissolved amount.

Which iodate concentration belongs in the solubility-product expression?

Use the post-precipitation excess concentration, 0.020 M, as the approximate equilibrium iodate concentration. Precipitation removes iodate ions, while mixing increases the solution volume; both changes must enter the calculation.

The stoichiometric ledger is:

Barium ions: initially 2 mmol; approximately zero after precipitation bookkeeping.
Iodate ions: initially 10 mmol; 4 mmol consumed; 6 mmol remaining.

n_{I O_{3}^{-}, l e f t} = 10 - 2 (2) = 6 m m o l

V_{t o t a l} = (200 + 100) m L = 300 m L = 0.300 L

Keep these three concentrations separate:

Stock: 0.10 M, before mixing.
Dilution only: accounts for the mixed volume but ignores consumption.

\frac{0.010}{0.300} = 0.0333 \dots M

Residual excess: accounts for both consumption and mixed volume.

c_{I O_{3}^{-}, e x c e s s} = \frac{0.006}{0.300} = 0.020 m o l L^{- 1}

Only the residual excess concentration supplies the common-ion background. The stock and dilution-only values both overstate the iodate left in solution.

How does the solubility product give the answer 3.95?

The official solution uses 0.020 M as the approximate equilibrium iodate concentration, giving a numerical entry of 3.95. It neglects the small iodate contribution from dissolution; the numerical check below justifies that approximation.

The dissolution equilibrium and solubility-product expression are:

B a (I O_{3})_{2} (s) ⇌ B a^{2 +} (a q) + 2 I O_{3}^{-} (a q)

K_{s p} = [B a^{2 +}] [I O_{3}^{-}]^{2}

The exponent two follows from the two iodate ions released per formula unit. Define the molar solubility and use the official common-ion approximation:

s = molar solubility, [B a^{2 +}] = s, [I O_{3}^{-}] \approx 0.020 M

Substitute and solve:

1.58 \times 10^{- 9} = s (0.020)^{2}

(0.020)^{2} = 4.0 \times 10^{- 4}

s = \frac{1.58 \times 10^{- 9}}{4.0 \times 10^{- 4}} = 3.95 \times 10^{- 6} M

M and moles per cubic decimetre are equivalent units. The question already supplies the power-of-ten multiplier, so enter 3.95, not the full solubility:

1 M = 1 m o l d m^{- 3}

$X = 3.95$ Check of the official approximation: the fuller iodate expression includes the ions released by dissolution. Using the calculated solubility:

[I O_{3}^{-}] = 0.020 + 2 s

2 s = 7.90 \times 10^{- 6} M

\frac{7.90 \times 10^{- 6}}{0.020} \times 100 = 0.0395 %

The added iodate is only 0.0395% of the residual excess concentration, so neglecting it is justified at the precision used here. This checks the official approximation, rather than replacing it with a cubic-equation solution; the retained answer is 3.95.

Why does accounting for dilution alone give the wrong answer?

Dilution alone overstates the iodate concentration because it ignores the iodate removed into the precipitate. This is a numerical-answer question, and the supplied record contains no answer options. The calculation below reconstructs a method error, not a supplied wrong option.

Using all 10 mmol of initial iodate in 300 mL gives:

[I O_{3}^{-}]_{w r o n g} = \frac{0.010}{0.300} = 0.0333 \dots M

Using that unrounded concentration:

s_{w r o n g} = \frac{1.58 \times 10^{- 9}}{(0.010 / 0.300)^{2}} = 1.422 \times 10^{- 6} M

$X_{w r o n g} = 1.422$ Dilution conserves the amount of a species only when no reaction consumes it. Here, precipitation removes 4 mmol of iodate.

The direction check agrees: overstating the common-ion concentration makes the calculated solubility too small. Use this sequence:

initial moles \to reaction consumption \to excess moles \to mixed-volume concentration \to K_{s p}

Which two practice questions check whether you understand the method?

Compare pure-water dissolution with dissolution in excess iodate: the ion concentrations need different expressions. These are original practice variations, not additional verified PYQs. Use the same solubility product in both:

K_{s p} = 1.58 \times 10^{- 9}

What is the solubility of barium iodate in pure water?

The worked answer is approximately:

s \approx 7.34 \times 10^{- 4} M

Original practice variation 1: Find the molar solubility of barium iodate in pure water, assuming ideal behaviour and no other equilibria. With no initial iodate, both ions come entirely from dissolution:

[B a^{2 +}] = s, [I O_{3}^{-}] = 2 s

K_{s p} = s (2 s)^{2} = 4 s^{3}

s = {(\frac{1.58 \times 10^{- 9}}{4})}^{1 / 3} \approx 7.34 \times 10^{- 4} M

This is much higher than in the original mixture. Excess iodate suppresses dissolution there:

s_{m i x t u r e} = 3.95 \times 10^{- 6} M

What happens if the residual excess iodate concentration is 0.040 M?

Doubling the excess iodate concentration reduces solubility to one-quarter under the same common-ion approximation. Original practice variation 2: Calculate the solubility at 0.040 M residual excess iodate using the same solubility product.

s = \frac{1.58 \times 10^{- 9}}{(0.040)^{2}} = 9.875 \times 10^{- 7} M

For pure water under the stated assumptions, use: $K_{s p} = 4 s^{3}$

For a residual excess iodate concentration that dominates the dissolution contribution, use:

c = residual excess iodate concentration, c ≫ 2 s, s \approx \frac{K_{s p}}{c^{2}}

Before substituting numbers, write where each dissolved ion comes from. Then select the matching expression.

Next step: the past-paper archive on JEEnius and search every JEE Main paper from 2002 and every Advanced paper from 2007, by year, subject or chapter, each with a worked solution (free).

For a worked example of the same idea, see Nuclear Fission and Q-Value JEE 2021: Energy Shares.

Frequently asked questions

What is the answer to the JEE Advanced 2025 barium iodate question?

The required numerical entry is 3.95. Using the residual iodate concentration of 0.020 M gives a molar solubility of 3.95 × 10^-6 mol dm^-3. Since the question asks for X in s = X × 10^-6 mol dm^-3, enter only 3.95.

Why is the iodate concentration 0.020 M after precipitation?

The mixture initially contains 2 mmol of barium ions and 10 mmol of iodate ions. Precipitating 2 mmol of barium iodate consumes 4 mmol of iodate, leaving 6 mmol in a total volume of 300 mL. The residual excess iodate concentration is therefore 0.006/0.300 = 0.020 M.

Why can we ignore 2s in the barium iodate solubility calculation?

The fuller equilibrium iodate concentration is 0.020 + 2s, where s is the molar solubility. Using s = 3.95 × 10^-6 M, the dissolution contribution 2s is only 0.0395% of 0.020 M. Neglecting it is justified for the approximation used in the official solution.

Why does using only dilution give the wrong solubility?

Dilution alone gives an iodate concentration of 0.010/0.300 = 0.0333... M, but precipitation consumes 4 mmol of the initial iodate. Using the dilution-only concentration in Ksp gives X = 1.422 instead of 3.95. Subtract reaction consumption before dividing the excess moles by the mixed volume.