(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.$K_{sp} = [Ba^{2+}][CO_3^{2-}]$Give

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(B) The precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product constant $(K_{sp})$.$K_{sp} = [Ba^{2+}][CO_3^{2-}]$Given $K_{sp} = 5.1 \times 10^{-9}$ and $[CO_3^{2-}] = 1.0 \times 10^{-4} \, M$.Substituting the values:$5.1 \times 10^{-9} = [Ba^{2+}] \times 1.0 \times 10^{-4}$$[Ba^{2+}] = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}}$$[Ba^{2+}] = 5.1 \times 10^{-5} \, M$

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Medium

Solid $Ba(NO_3)_2$ is gradually dissolved in $1.0 \times 10^{-4} \, M \, Na_2CO_3$ solution. At which concentration of $Ba^{2+}$ will the precipitation begin? $(K_{sp} \, BaCO_3 = 5.1 \times 10^{-9})$

A
$4.1 \times 10^{-5} \, M$
B
$5.1 \times 10^{-5} \, M$
C
$8.1 \times 10^{-8} \, M$
D
$8.1 \times 10^{-7} \, M$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solubility product

Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

MediumAIIMS 2016

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The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

Medium

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$A$ salt $MX$ has $K_{sp} = 4 \times 10^{-10}$. What value of $K_{sp}$ must another salt $MX_3$ have if the molar solubility of the two salts is to be identical?

Medium

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The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$

DifficultNEET 2026

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Which of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions?
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.

DifficultJEE MAIN 2025

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Salt	$K_{sp}$ at $298 K$
$AgBr$	$5.0 \times 10^{-13}$
$Zn(OH)_2$	$1.0 \times 10^{-15}$
$Hg_2Cl_2$	$1.3 \times 10^{-18}$

Solid $Ba(NO_3)_2$ is gradually dissolved in $1.0 \times 10^{-4} \, M \, Na_2CO_3$ solution. At which concentration of $Ba^{2+}$ will the precipitation begin? $(K_{sp} \, BaCO_3 = 5.1 \times 10^{-9})$

Similar Questions

Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

The solubility product of $AgCl$ is $1.5625 \times 10^{-10}$ at $25\ ^oC$. Its solubility in grams per litre will be

$A$ salt $MX$ has $K_{sp} = 4 \times 10^{-10}$. What value of $K_{sp}$ must another salt $MX_3$ have if the molar solubility of the two salts is to be identical?

The correct order of solubility of the given salts in water at $298 K$ is:Salt$K_{sp}$ at $298 K$$AgBr$$5.0 \times 10^{-13}$$Zn(OH)_2$$1.0 \times 10^{-15}$$Hg_2Cl_2$$1.3 \times 10^{-18}$

Which of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions? Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.

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Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

The correct order of solubility of the given salts in water at $298 K$ is:
Salt $K_{sp}$ at $298 K$
$AgBr$ $5.0 \times 10^{-13}$
$Zn(OH)_2$ $1.0 \times 10^{-15}$
$Hg_2Cl_2$ $1.3 \times 10^{-18}$

Which of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions?
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.