(A) Given concentration of $Na_2CO_3 = 1.0 \times 10^{-4} \ M$.Since $Na_2CO_3$ is a strong electrolyte,$[CO_3^{2-}] = 1.0 \times 10^{-4} \ M$.Precipi

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(A) Given concentration of $Na_2CO_3 = 1.0 \times 10^{-4} \ M$.Since $Na_2CO_3$ is a strong electrolyte,$[CO_3^{2-}] = 1.0 \times 10^{-4} \ M$.Precipitation of $BaCO_3$ begins when the ionic product exceeds the solubility product $(K_{sp})$.At the point of precipitation,$[Ba^{2+}][CO_3^{2-}] = K_{sp} = 5.1 \times 10^{-9}$.Therefore,$[Ba^{2+}] = \frac{K_{sp}}{[CO_3^{2-}]} = \frac{5.1 \times 10^{-9}}{1.0 \times 10^{-4}} = 5.1 \times 10^{-5} \ M$.

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

Medium

Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \ M \ Na_2CO_3$ solution. At which concentration of $Ba^{2+}$,precipitate of $BaCO_3$ begins to form? ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

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A
$5.1 \times 10^{-5} \ M$
B
$7.1 \times 10^{-8} \ M$
C
$4.1 \times 10^{-5} \ M$
D
$8.1 \times 10^{-7} \ M$

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6-2.Equilibrium-II (Ionic Equilibrium)

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Solid $Ba(NO_3)_2$ is gradually dissolved in a $1.0 \times 10^{-4} \ M \ Na_2CO_3$ solution. At which concentration of $Ba^{2+}$,precipitate of $BaCO_3$ begins to form? ($K_{sp}$ for $BaCO_3 = 5.1 \times 10^{-9}$)

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The solubility product of $NiS$ is $4.9 \times 10^{-5}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

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The $K_{sp}$ for the following dissociation is $1.6 \times 10^{-5}$:
$PbCl_{2(s)} \rightleftharpoons Pb^{2+}_{(aq)} + 2Cl^-_{(aq)}$
Which of the following choices is correct for a mixture of $300 \ mL$ of $0.134 \ M \ Pb(NO_3)_2$ and $100 \ mL$ of $0.4 \ M \ NaCl$?