Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Ionic Equilibrium question

2025 · Shift 2 · Q10
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Advanced
  3. /Chemistry
  4. /Ionic Equilibrium
  5. /2025 · Shift 2 · Q10

Ionic Equilibrium question

2025 · Shift 2 · Q10

JEE AdvancedChemistryIonic EquilibriumNumerical+4 / −1
The solubility of barium iodate in an aqueous solution prepared by mixing 200 mL of 0.010 M barium nitrate with 100 mL of 0.10 M sodium iodate is X×10−6 moldm−3\boldsymbol{X} \times 10^{-6} \mathrm{~mol} \mathrm{dm}^{-3}X×10−6 moldm−3. The value of X\boldsymbol{X}X is ‾\underline{\hspace{2cm}}​. Use: Solubility product constant (Ksp)\left(K_{\mathrm{sp}}\right)(Ksp​) of barium iodate =1.58×10−9=1.58 \times 10^{-9}=1.58×10−9
Numerical answer
View written solutionFree

Correct answer: 3.85TO4.15

  1. Write the dissolution equilibrium

For barium iodate:

Ba(IO3)2(s)⇌Ba2++2 IO3−\mathrm{Ba(IO_3)_2(s)} \rightleftharpoons \mathrm{Ba^{2+}} + 2\,\mathrm{IO_3^-}Ba(IO3​)2​(s)⇌Ba2++2IO3−​

Hence,

Ksp=[Ba2+][IO3−]2K_{sp} = [\mathrm{Ba^{2+}}][\mathrm{IO_3^-}]^2Ksp​=[Ba2+][IO3−​]2

Given:

Ksp=1.58×10−9K_{sp} = 1.58 \times 10^{-9}Ksp​=1.58×10−9
  1. Find concentrations after mixing

We mix:

  • 200 mL200\ \text{mL}200 mL of 0.010 M Ba(NO3)20.010\ \text{M } \mathrm{Ba(NO_3)_2}0.010 M Ba(NO3​)2​
  • 100 mL100\ \text{mL}100 mL of 0.10 M NaIO30.10\ \text{M } \mathrm{NaIO_3}0.10 M NaIO3​

Total volume after mixing:

Vtotal=200+100=300 mL=0.300 LV_{\text{total}} = 200 + 100 = 300\ \text{mL} = 0.300\ \text{L}Vtotal​=200+100=300 mL=0.300 L

Moles of Ba2+\mathrm{Ba^{2+}}Ba2+

nBa2+=0.010×0.200=2.0×10−3 moln_{\mathrm{Ba^{2+}}} = 0.010 \times 0.200 = 2.0 \times 10^{-3}\ \text{mol}nBa2+​=0.010×0.200=2.0×10−3 mol

So concentration after mixing is:

[Ba2+]=2.0×10−30.300=6.67×10−3 M[\mathrm{Ba^{2+}}] = \frac{2.0 \times 10^{-3}}{0.300} = 6.67 \times 10^{-3}\ \text{M}[Ba2+]=0.3002.0×10−3​=6.67×10−3 M

Moles of IO3−\mathrm{IO_3^-}IO3−​

nIO3−=0.10×0.100=1.0×10−2 moln_{\mathrm{IO_3^-}} = 0.10 \times 0.100 = 1.0 \times 10^{-2}\ \text{mol}nIO3−​​=0.10×0.100=1.0×10−2 mol

So concentration after mixing is:

[IO3−]=1.0×10−20.300=3.33×10−2 M[\mathrm{IO_3^-}] = \frac{1.0 \times 10^{-2}}{0.300} = 3.33 \times 10^{-2}\ \text{M}[IO3−​]=0.3001.0×10−2​=3.33×10−2 M
  1. Let the solubility in this mixed solution be sss

If additional barium iodate dissolves by s mol dm−3s\ \text{mol dm}^{-3}s mol dm−3, then:

[Ba2+]=6.67×10−3+s[\mathrm{Ba^{2+}}] = 6.67 \times 10^{-3} + s[Ba2+]=6.67×10−3+s [IO3−]=3.33×10−2+2s[\mathrm{IO_3^-}] = 3.33 \times 10^{-2} + 2s[IO3−​]=3.33×10−2+2s

Since the solution already contains large common-ion concentrations, sss will be very small. So:

6.67×10−3+s≈6.67×10−36.67 \times 10^{-3} + s \approx 6.67 \times 10^{-3}6.67×10−3+s≈6.67×10−3 3.33×10−2+2s≈3.33×10−23.33 \times 10^{-2} + 2s \approx 3.33 \times 10^{-2}3.33×10−2+2s≈3.33×10−2

Thus,

Ksp≈(6.67×10−3)(3.33×10−2)2K_{sp} \approx (6.67 \times 10^{-3})(3.33 \times 10^{-2})^2Ksp​≈(6.67×10−3)(3.33×10−2)2

But this is the ionic product of the mixed solution before considering dissolution/precipitation:

Q=(6.67×10−3)(3.33×10−2)2approx7.4×10−6Q = (6.67 \times 10^{-3})(3.33 \times 10^{-2})^2 approx 7.4 \times 10^{-6}Q=(6.67×10−3)(3.33×10−2)2approx7.4×10−6

Since

Q≫KspQ \gg K_{sp}Q≫Ksp​

the mixed solution is supersaturated, so in fact precipitation occurs first.

Therefore, we must calculate the equilibrium concentrations after precipitation, and then find the solubility of barium iodate in that equilibrium mixture.

  1. Precipitation stoichiometry

Reaction:

Ba2++2IO3−→Ba(IO3)2(s)\mathrm{Ba^{2+}} + 2\mathrm{IO_3^-} \to \mathrm{Ba(IO_3)_2(s)}Ba2++2IO3−​→Ba(IO3​)2​(s)

Initial moles:

Ba2+=2.0×10−3 mol\mathrm{Ba^{2+}} = 2.0 \times 10^{-3}\ \text{mol}Ba2+=2.0×10−3 mol IO3−=1.0×10−2 mol\mathrm{IO_3^-} = 1.0 \times 10^{-2}\ \text{mol}IO3−​=1.0×10−2 mol

To precipitate all Ba2+\mathrm{Ba^{2+}}Ba2+, required IO3−\mathrm{IO_3^-}IO3−​ is:

2×2.0×10−3=4.0×10−3 mol2 \times 2.0 \times 10^{-3} = 4.0 \times 10^{-3}\ \text{mol}2×2.0×10−3=4.0×10−3 mol

Available iodate is 1.0×10−21.0 \times 10^{-2}1.0×10−2 mol, so IO3−\mathrm{IO_3^-}IO3−​ is in excess and almost all Ba2+\mathrm{Ba^{2+}}Ba2+ precipitates.

Moles of iodate left after complete precipitation of barium:

1.0×10−2−4.0×10−3=6.0×10−3 mol1.0 \times 10^{-2} - 4.0 \times 10^{-3} = 6.0 \times 10^{-3}\ \text{mol}1.0×10−2−4.0×10−3=6.0×10−3 mol

Thus, after precipitation, excess iodate concentration is:

[IO3−]≈6.0×10−30.300=2.0×10−2 M[\mathrm{IO_3^-}] \approx \frac{6.0 \times 10^{-3}}{0.300} = 2.0 \times 10^{-2}\ \text{M}[IO3−​]≈0.3006.0×10−3​=2.0×10−2 M

Now the precipitated barium iodate is present as a solid, and its solubility in this iodate-containing solution is to be found.

  1. Calculate solubility in presence of common ion

Let solubility of Ba(IO3)2\mathrm{Ba(IO_3)_2}Ba(IO3​)2​ in this final solution be s mol dm−3s\ \text{mol dm}^{-3}s mol dm−3.

Then:

[Ba2+]=s[\mathrm{Ba^{2+}}] = s[Ba2+]=s [IO3−]=2.0×10−2+2s[\mathrm{IO_3^-}] = 2.0 \times 10^{-2} + 2s[IO3−​]=2.0×10−2+2s

Since sss is very small,

[IO3−]≈2.0×10−2[\mathrm{IO_3^-}] \approx 2.0 \times 10^{-2}[IO3−​]≈2.0×10−2

Using KspK_{sp}Ksp​:

Ksp=s(2.0×10−2)2K_{sp} = s(2.0 \times 10^{-2})^2Ksp​=s(2.0×10−2)2 s=1.58×10−9(2.0×10−2)2s = \frac{1.58 \times 10^{-9}}{(2.0 \times 10^{-2})^2}s=(2.0×10−2)21.58×10−9​ s=1.58×10−94.0×10−4=3.95×10−6 mol dm−3s = \frac{1.58 \times 10^{-9}}{4.0 \times 10^{-4}} = 3.95 \times 10^{-6}\ \text{mol dm}^{-3}s=4.0×10−41.58×10−9​=3.95×10−6 mol dm−3

So,

X=3.95X = 3.95X=3.95
  1. Comparison with stored correct answer

Stored correct answer range: 3.853.853.85 to 4.154.154.15

Our value:

X=3.95X = 3.95X=3.95

This lies within the given range, so the answer agrees.

PreviousNext

More from Ionic Equilibrium

  • On decreasing the pH from 7 to 2 , the solubility of a sparingly soluble salt (MX) of a weak acid (HX) increased from 10−4 mol L−1 to 10−3 mol L−1. The p Ka​…2023 · MCQ
  • A solution is prepared by mixing 0.01 mol each of H2​CO3​,NaHCO3​,Na2​CO3​, and NaOH in 100 mL of water. pH of the resulting…2022 · Numerical
  • Concentration of H2​SO4​ and Na2​SO4​ in a solution is 1M and 1.8×10−2M, respectively. Molar solubility of PbSO4​ in the same solution is X×10−YM…2022 · Numerical
  • A solution of 0.1 M weak base (B) is titrated with 0.1 M of a strong acid (HA). The variation of pH of the solution with the volume of HA added is shown in the figure below. What is the pKb of the base? The neutralisation reaction is given… Includes diagram2020 · Numerical
  • An acidified solution of 0.05 M Zn2+ is saturated with 0.1 M H2​S. What is the minimum molar concentration (M) of H+ required to prevent the precipitation of ZnS? Use Ksp(ZnS) = 1.25 × 10 − 22 and overall…2020 · Numerical
  • The solubility of a salt of weak acid (AB) at pH3 is Y×10−3molL−1. The value of Y is ​. (Given that the value of solubility product of AB (Ksp​)=2×10−10…2018 · Numerical
  • Dilution processes of different aqueous solutions, with water, are given in LIST - I. The effects of dilution of the solutions on [H+] are given in LIST - II (Note: Degree of dissociation (a) of weak acid and weak… Includes table2018 · MCQ
  • Paragraph When 100 mL of 1.0 M KCl was mixed with 100 mL of 1.0 M NaOH in an insulated beaker at constant pressure, a temperature increase of 5.7o C was measured for the beaker and its contents (Expt. 1). Because the enthalpy of…2015 · Multiple correct