The solubility of AgCl is 1 times 10^{-5} mo | vedclass

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

Question

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as: $K_{sp} = (	ext{solubility})^2 = (1 	imes 10^{-5})^2 = 1 	imes 10^{-10}$

Vedclass · Accepted Answer

$1 	imes 10^{-9}$

Vedclass · Answer

(C) The solubility product constant $(K_{sp})$ of $AgCl$ is calculated as: $K_{sp} = (	ext{solubility})^2 = (1 	imes 10^{-5})^2 = 1 	imes 10^{-10}$. Let the solubility of $AgCl$ in $0.1 \ M \ NaCl$ be $x \ mol/L$. In the presence of $0.1 \ M \ NaCl$,the concentration of chloride ions is $[Cl^-] = (x + 0.1) \ M$. Since $x$ is very small compared to $0.1$,we can approximate $[Cl^-] \approx 0.1 \ M$. The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$. Substituting the values: $1 	imes 10^{-10} = x 	imes 0.1$. Solving for $x$: $x = \frac{1 	imes 10^{-10}}{0.1} = 1 	imes 10^{-9} \ mol/L$.

The solubility of $AgCl$ is $1 \times 10^{-5} \ mol/L$. Its solubility in $0.1 \ M$ sodium chloride solution is

Similar Questions

The solubility product of $Mg(OH)_2$ is $1 \times 10^{-12}$. At what $pH$ will precipitation occur in a $0.01 \ M$ $MgCl_2$ solution?

The solubility of $A_2X_3$ is $y \ mol \ m^{-3}$. Its solubility product is equal to .......

Which of the following happens when $NH_{4}OH$ is added gradually to the solution containing $1 \ M \ A^{2+}$ and $1 \ M \ B^{3+}$ ions?
Given: $K_{sp}[A(OH)_{2}] = 9 \times 10^{-10}$ and $K_{sp}[B(OH)_{3}] = 27 \times 10^{-18}$ at $298 \ K$.

The concentration of a saturated solution of $Mg(OH)_2$ is $8.2 \times 10^{-4} \ g \ L^{-1}$ at $298 \ K$. Calculate its solubility product $(K_{sp})$. (Molar mass of $Mg(OH)_2 = 58 \ g \ mol^{-1}$)

If $K_{sp}$ of $Ag_2CO_3$ is $8 \times 10^{-12},$ the molar solubility of $Ag_2CO_3$ in $0.1 \ M \ AgNO_3$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module