The solubility of $AgCl$ is $1.435 \times {10^{ - 5}}\,g{L^{ - 1}}$ at $30^{°}$ $C$ temp. then calculate its solubility product.
$1.0 \times 10^{-14}$
${K_{sp}}$ for $Cr{(OH)_3}$ is $2.7 \times {10^{ – 31}}$. What is its solubility in moles / litre.
The required amount of $KBr$ (molar mass $= 119$) in gram to start the precipitation of $AgBr$ in $500\, mL$ solution of $0.05\, M\, AgNO_3$ will be ($K_{SP}$ of $AgBr = 5 \times 10^{-13}$)
The molar solubility of $PbI_2$ in $0.2\,M\,\, Pb(NO_3)_2$ solution in terms of solubility product of $PbI_2$ is
How many grams of $Ca{C_2}{O_4}$ will dissolve in distilled water to make one litre of saturated solution ? (Solubility product of $Ca{C_2}{O_4}$ is $2.5 \times {10^{ – 9}}\,mol{e^2}\,litr{e^{ – 2}}$ and its molecular weight is $128$)
Solubility of $M{X_2}$ type electrolyte is $0.5 \times {10^{ – 4}}\,mole/litre$. The value of ${K_{sp}}$ of the electrolyte is
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