Question

Excess of Ag2CrO4 was dissolved in distilled water its solubility was found to be 1.3×10^-4 mol/dm3. what is the solubility product?

Options are given above

Answer 1

Answer: $K_{sp}=[2.6\times 10^{-4}]^2[1.3\times 10^{-4}]$

Explanation:

Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as $K_{sp}$

The equation for the ionization of the silver chromate is given as:

$Ag_2CrO_4\leftrightharpoons 2Ag^++CrO_4^{2-}$

We are given:

Solubility of $Ag_2CrO_4$ =$1.3\times 10^{-4}$ mol/L

By stoichiometry of the reaction:

1 mole of $Ag_2CrO_4$ gives 2 moles of $Ag^{+}$ and 1 mole of $CrO_4^{2-}$.

Expression for the equilibrium constant of $Ag_2CrO_4$ will be:

$K_{sp}=[Ag^+]^2[CrO_4^{2-}]$

$K_{sp}=[2\times 1.3\times 10^{-4}]^2[1.3\times 10^{-4}]$

$K_{sp}=[2.6\times 10^{-4}]^2[1.3\times 10^{-4}]$

Hence, the solubility product of $Ag_2CrO_4$ is  $K_{sp}=[2.6\times 10^{-4}]^2[1.3\times 10^{-4}]$

Answer 2

Answer:

AgCro4

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