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

If the concentration of $Ag^{+}$ ions in the saturated solution of $Ag_2CO_3$ is $1.20 \times 10^{-4} \ mol \ L^{-1}$,then find the solubility product of $Ag_2CO_3$.

The $K_{sp}$ for $BaSO_4$ at $25\ ^\circ C$ 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 (in $M$):

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

$A$ solution is $0.1 \ M$ in $Cl^{-}$ and $0.001 \ M$ in $CrO_{4}^{2-}$. Solid $AgNO_{3}$ is gradually added to it. Assuming that the addition does not change in volume and $K_{sp}(AgCl) = 1.7 \times 10^{-10} \ M^{2}$ and $K_{sp}(Ag_{2}CrO_{4}) = 1.9 \times 10^{-12} \ M^{3}$. Select the correct statement from the following:

Assertion : $BaCO_3$ is more soluble in $HNO_3$ than in plain water.
Reason : Carbonate is a weak base and reacts with the $H^+$ from the strong acid,causing the barium salt to dissociate.