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Ionic Equilibrium question

2009 · Shift 0 · Q6
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Ionic Equilibrium question

2009 · Shift 0 · Q6

JEE MainChemistryIonic EquilibriumMCQ+4 / −1
Solid Ba(NO3)2Ba(NO_3)_2Ba(NO3​)2​ is gradually dissolved in a 1.0 ×\times× 10-4 M Na2CO3Na_2CO_3Na2​CO3​ solution. At what concentration of Ba2+Ba^{2+}Ba2+ will a precipitate begin to form ? (Ksp for BaCO3BaCO_3BaCO3​ = 5.1 ×\times× 10−9 )
  1. A
    5.1 ×\times× 10-5 M
  2. B
    8.1 ×\times× 10-8 M
  3. C
    8.1 ×\times× 10-7 M
  4. D
    4.1 ×\times× 10-5 M
View written solutionFree

Correct answer: A

  1. Condition for precipitation

A precipitate of BaCO3BaCO_3BaCO3​ begins to form when the ionic product becomes equal to the solubility product:

[Ba2+][CO32−]=Ksp[Ba^{2+}][CO_3^{2-}] = K_{sp}[Ba2+][CO32−​]=Ksp​

Given:

Ksp(BaCO3)=5.1×10−9K_{sp}(BaCO_3)=5.1\times 10^{-9}Ksp​(BaCO3​)=5.1×10−9

  1. Concentration of carbonate ion

The solution is 1.0×10−4 M1.0\times 10^{-4}\,M1.0×10−4M Na2CO3Na_2CO_3Na2​CO3​. Since Na2CO3Na_2CO_3Na2​CO3​ is a strong electrolyte, initially:

[CO32−]=1.0×10−4 M[CO_3^{2-}] = 1.0\times 10^{-4}\,M[CO32−​]=1.0×10−4M

  1. Find the minimum [Ba2+][Ba^{2+}][Ba2+] needed for precipitation

At the point where precipitation just starts:

[Ba2+]=Ksp[CO32−][Ba^{2+}] = \frac{K_{sp}}{[CO_3^{2-}]}[Ba2+]=[CO32−​]Ksp​​

So,

[Ba2+]=5.1×10−91.0×10−4[Ba^{2+}] = \frac{5.1\times 10^{-9}}{1.0\times 10^{-4}}[Ba2+]=1.0×10−45.1×10−9​

[Ba2+]=5.1×10−5 M[Ba^{2+}] = 5.1\times 10^{-5}\,M[Ba2+]=5.1×10−5M

  1. Match with options

This corresponds to:

A: 5.1×10−5 M5.1 \times 10^{-5}\,M5.1×10−5M

  1. Comparison with stored correct answer

Stored correct answer: A

Derived answer: A

They match.

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