The solubility product constants (K_{sp}) for | vedclass

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

Question

(A) For $1:1$ electrolytes like $AgCl$,$AgBr$,and $AgI$,the solubility $S = \sqrt{K_{sp}}$.For $AgCl$: $S = \sqrt{2 	imes 10^{-10}} \approx 1.41 	im

Vedclass · Accepted Answer

$Ag_2CO_3$

Vedclass · Answer

(A) For $1:1$ electrolytes like $AgCl$,$AgBr$,and $AgI$,the solubility $S = \sqrt{K_{sp}}$.For $AgCl$: $S = \sqrt{2 	imes 10^{-10}} \approx 1.41 	imes 10^{-5} \ M$.For $AgBr$: $S = \sqrt{5 	imes 10^{-13}} \approx 7.07 	imes 10^{-7} \ M$.For $AgI$: $S = \sqrt{8 	imes 10^{-17}} \approx 8.94 	imes 10^{-9} \ M$.For $Ag_2CO_3$ ($1:2$ electrolyte),$K_{sp} = 4S^3$,so $S = (K_{sp}/4)^{1/3} = (8 	imes 10^{-12} / 4)^{1/3} = (2 	imes 10^{-12})^{1/3} \approx 1.26 	imes 10^{-4} \ M$.Comparing the values,$Ag_2CO_3$ has the highest solubility.

The solubility product constants $(K_{sp})$ for some silver salts are given as: $AgCl = 2 \times 10^{-10}$,$AgBr = 5 \times 10^{-13}$,$Ag_2CO_3 = 8 \times 10^{-12}$,and $AgI = 8 \times 10^{-17}$. Which of the following salts has the highest solubility?

Similar Questions

$K_{sp}$ for $Cr(OH)_3$ is $2.7 \times 10^{-31}$. What is its solubility in moles / litre?

On the addition of a solution containing $CrO_4^{2-}$ ions to a solution containing $Ba^{2+}$,$Sr^{2+}$,and $Ca^{2+}$ ions,the precipitate obtained first will be of

What is the concentration of $Ba^{2+}$ when $BaF_2$ $(K_{sp} = 1.0 \times 10^{-6})$ begins to precipitate from a solution that is $0.30 \, M \ F^{-}$?

If the solubility of calcium hydroxide is $\sqrt{3}$,then its solubility product will be

Solubility product constant $(K_{sp})$ of salts of types $MX$,$MX_2$ and $M_3X$ at temperature '$T$' are $4.0 \times 10^{-8}$,$3.2 \times 10^{-14}$ and $2.7 \times 10^{-15}$,respectively. Solubilities $(\text{mol } dm^{-3})$ of the salts at temperature '$T$' are in the order:

Vedclass Test Series

Exam Paper Generator

Online Exam Module