The solubility of AgCl_{(s)} with solubility | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(B) The dissociation of $AgCl$ is given by: $AgCl_{(s)} \rightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$The solubility product expression is $K_{sp}

Vedclass · Accepted Answer

$1.6 	imes 10^{-9} \ M$

Vedclass · Answer

(B) The dissociation of $AgCl$ is given by: $AgCl_{(s)} ightleftharpoons Ag^{+}_{(aq)} + Cl^{-}_{(aq)}$The solubility product expression is $K_{sp} = [Ag^{+}][Cl^{-}]$.In a $0.1 \ M \ NaCl$ solution,the concentration of $Cl^{-}$ ions is dominated by the $NaCl$ dissociation,so $[Cl^{-}] \approx 0.1 \ M$.Let $S$ be the solubility of $AgCl$ in the presence of $NaCl$. Then $[Ag^{+}] = S$.Substituting the values into the $K_{sp}$ expression:$1.6 	imes 10^{-10} = S 	imes 0.1$Solving for $S$:$S = \frac{1.6 	imes 10^{-10}}{0.1} = 1.6 	imes 10^{-9} \ M$.

The solubility of $AgCl_{(s)}$ with solubility product $1.6 \times 10^{-10}$ in $0.1 \ M \ NaCl$ solution would be

Similar Questions

Which hydroxide will have the lowest value of solubility product at normal temperature $({25\,^oC})$?

If the solubility of $PbBr_2$ is $S \, mol/L$,and it undergoes $100\%$ ionization,then the solubility product constant $(K_{sp})$ is equal to:

The precipitation occurs if the ionic product is

The $K_{sp}$ of $AgBr$ at $25 \, ^\circ C$ is $4.9 \times 10^{-13}$. If the molar mass of $AgBr$ is $188 \, g/mol$,how much $AgBr$ is required to saturate $20 \, L$ of water?

Which of the following is least soluble in water?

Vedclass Test Series

Exam Paper Generator

Online Exam Module