(A) $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$ $0.1 \ M \quad \quad \quad 0.2 \ M \quad 0.1 \ M$ For $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$: $K_{

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(A) $K_2SO_4 \rightarrow 2K^+ + SO_4^{2-}$ $0.1 \ M \quad \quad \quad 0.2 \ M \quad 0.1 \ M$ For $BaSO_4 \rightleftharpoons Ba^{2+} + SO_4^{2-}$: $K_{sp} = [Ba^{2+}][SO_4^{2-}]$ $1 \times 10^{-10} = S \times (S + 0.1)$ Since $S$ is very small,$S + 0.1 \approx 0.1$. $1 \times 10^{-10} = S \times 0.1$ $S = 10^{-9} \ mol \ L^{-1}$ Solubility in $g \ L^{-1} = S \times \text{Molar mass} = 10^{-9} \times 233 \ g \ L^{-1} = 233 \times 10^{-9} \ g \ L^{-1}$. Thus,the value is $233$.

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

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The solubility product of $BaSO_4$ is $1 \times 10^{-10}$ at $298 \ K$. The solubility of $BaSO_4$ in $0.1 \ M \ K_2SO_4 \ (aq)$ solution is $......... \times 10^{-9} \ g \ L^{-1}$ (nearest integer). Given: Molar mass of $BaSO_4$ is $233 \ g \ mol^{-1}$.

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A
$233$
B
$232$
C
$231$
D
$234$

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The solubility product of $BaSO_4$ is $1 \times 10^{-10}$ at $298 \ K$. The solubility of $BaSO_4$ in $0.1 \ M \ K_2SO_4 \ (aq)$ solution is $......... \times 10^{-9} \ g \ L^{-1}$ (nearest integer). Given: Molar mass of $BaSO_4$ is $233 \ g \ mol^{-1}$.

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Solid $Ba(NO_3)_2$ is gradually added to a $1.0 \times 10^{-4} \, M \, Na_2CO_3$ solution. At what concentration of $Ba^{2+}$ will precipitation start? (Given $K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

The expression for solubility product of mercurous iodide $(Hg_2I_2)$ is

Among different types of sparingly soluble salts having nearly the same solubility product constants $(K_{sp})$,the least soluble salt is that which:

The solubility of $BaSO_4$ in water is $2.33 \times 10^{-3} \ g/L$. Its solubility product will be (molecular weight of $BaSO_4 = 233$).

Assign $A, B, C, D$ from the given type of reaction.$HgS(s) + Na_2S(aq) \rightleftharpoons Na_2[HgS_2](aq)$

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Assign $A, B, C, D$ from the given type of reaction.
$HgS(s) + Na_2S(aq) \rightleftharpoons Na_2[HgS_2](aq)$