The solubility product of the sparingly soluble salt $AB_2$ is $2.56 \times 10^{-10}$ at $298 \ K$. Calculate its solubility in $mol \ dm^{-3}$ at the same temperature?

The concentration of sulphide ion in $0.1 \ M \ HCl$ solution saturated with hydrogen sulphide is $1.0 \times 10^{-19} \ M$. If $10 \ mL$ of this is added to $5 \ mL$ of $0.04 \ M$ solution of the following: $FeSO_4, MnCl_2, ZnCl_2$ and $CdCl_2$,in which of these solutions will precipitation take place? Given $K_{sp}$ for $FeS = 6.3 \times 10^{-18}, MnS = 2.5 \times 10^{-13}, ZnS = 1.6 \times 10^{-24}, CdS = 8.0 \times 10^{-27}$.

Assign $A$,$B$,$C$,$D$ from the given type of reaction.
$Ag_2CO_3 \downarrow + 2HCl \longrightarrow 2AgCl \downarrow + CO_2 \uparrow + H_2O$

The solubility of an electrolyte of type $MX_2$ is $0.5 \times 10^{-4} \ mol/L$. Find the $K_{sp}$ of the electrolyte.

Assign $A, B, C, D$ from the given type of reaction.
$2AgNO_3 + Na_2C_2O_4 \longrightarrow Ag_2C_2O_4 \downarrow + 2NaNO_3$

$K_{sp}$ for $BaSO_4$ at $25 \, ^oC$ is $1.1 \times 10^{-10}$. To make the new solubility equal to $1.1 \times 10^{-8} \, M$,it is necessary to use a solution of $Na_2SO_4$ of the following concentration $....... \ M$