The K_{sp} of AgCl is 1.8 times 10^{-10}. The | vedclass

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

Question

(A) The precipitation of $AgCl$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.For the reaction $AgCl(s) \rightlefth

Vedclass · Accepted Answer

$4.5 	imes 10^{-8} \ M$

Vedclass · Answer

(A) The precipitation of $AgCl$ occurs when the ionic product exceeds the solubility product constant $(K_{sp})$.For the reaction $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$,the solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.Given $K_{sp} = 1.8 	imes 10^{-10}$ and $[Ag^+] = 4 	imes 10^{-3} \ M$.To find the minimum concentration of $Cl^-$ required for precipitation,we set the ionic product equal to $K_{sp}$:$[Cl^-] = \frac{K_{sp}}{[Ag^+]} = \frac{1.8 	imes 10^{-10}}{4 	imes 10^{-3}} = 0.45 	imes 10^{-7} \ M = 4.5 	imes 10^{-8} \ M$.

The $K_{sp}$ of $AgCl$ is $1.8 \times 10^{-10}$. The concentration of $Cl^-$ required to initiate the precipitation of $AgCl$ in a solution containing $4 \times 10^{-3} \ M \ Ag^+$ is:

Similar Questions

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

Explain the relationship between solubility $(S)$ and solubility product constant $(K_{sp})$.

What is the $pH$ at which $Mg(OH)_2$ starts to precipitate from a solution containing $0.1 \ M \ Mg^{2+}$ ions? (Given $K_{sp}$ for $Mg(OH)_2 = 1.0 \times 10^{-11}$)

The solubility product of lead bromide is $8 \times 10^{-5}$. If the salt is $80\%$ dissociated in a saturated solution,find the solubility of the salt.

What is the molar solubility of $Al(OH)_3$ in $0.2 \ M \ NaOH$ solution? Given that,the solubility product of $Al(OH)_3 = 2.4 \times 10^{-24}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module