The solubility product of AgCl is 4.0 times 1 | vedclass

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

Question

(C) The dissociation of $AgCl$ is given by: $AgCl(s) \rightleftharpoons Ag^+(aq) + Cl^-(aq)$.$CaCl_2$ is a strong electrolyte and dissociates complete

Vedclass · Accepted Answer

$5.0 	imes 10^{-9} \ m$

Vedclass · Answer

(C) The dissociation of $AgCl$ is given by: $AgCl(s) ightleftharpoons Ag^+(aq) + Cl^-(aq)$.$CaCl_2$ is a strong electrolyte and dissociates completely as: $CaCl_2(aq) ightarrow Ca^{2+}(aq) + 2Cl^-(aq)$.Since the concentration of $CaCl_2$ is $0.04 \ m$,the concentration of $Cl^-$ ions from $CaCl_2$ is $2 	imes 0.04 \ m = 0.08 \ m$.Let the solubility of $AgCl$ be $s$. Then $[Ag^+] = s$ and $[Cl^-] = (0.08 + s) \approx 0.08 \ m$ (since $s$ is very small).The solubility product expression is $K_{sp} = [Ag^+][Cl^-]$.$4.0 	imes 10^{-10} = s 	imes 0.08$.$s = \frac{4.0 	imes 10^{-10}}{0.08} = 5.0 	imes 10^{-9} \ m$.

The solubility product of $AgCl$ is $4.0 \times 10^{-10}$ at $298 \ K$. The solubility of $AgCl$ in $0.04 \ m \ CaCl_2$ will be

Similar Questions

The value of $K_{sp}$ for $B_2CO_3$ is $3.2 \times 10^{-5}$. The concentration of $B^{+}$ in its saturated solution is :-

At $25^{\circ} C$, the solubility product of a salt of $MX_{2}$ type is $3.2 \times 10^{-8}$ in water. The solubility (in $mol / L$) of $MX_{2}$ in water at the same temperature will be:

$1 \ dm^{3}$ solution containing $10^{-5} \ mol$ each of $Cl^{-}$ ions and $CrO_{4}^{2-}$ ions is treated with $10^{-4} \ mol$ of silver nitrate. Which one of the following observations is made?
$[K_{sp} \ Ag_{2}CrO_{4} = 4 \times 10^{-12}]$
$[K_{sp} \ AgCl = 1 \times 10^{-10}]$

The solubility of a salt $A_2B_3$ is $10^{-3} \ M$. What will be its solubility product?

The solubility of an $MX_2$ type electrolyte is $0.5 \times 10^{-4} \ mol/L$. The value of $K_{sp}$ of the electrolyte is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module