The solubility products (K_{sp}) of three spa | vedclass

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

Question

(C) For a salt of type $M_2X$,the solubility product is $K_{sp} = (2s)^2(s) = 4s^3$,so $s = (K_{sp}/4)^{1/3}$.For a salt of type $MX$,the solubility p

Vedclass · Accepted Answer

$MX > M_2X > MX_3$

Vedclass · Answer

(C) For a salt of type $M_2X$,the solubility product is $K_{sp} = (2s)^2(s) = 4s^3$,so $s = (K_{sp}/4)^{1/3}$.For a salt of type $MX$,the solubility product is $K_{sp} = (s)(s) = s^2$,so $s = (K_{sp})^{1/2}$.For a salt of type $MX_3$,the solubility product is $K_{sp} = (s)(3s)^3 = 27s^4$,so $s = (K_{sp}/27)^{1/4}$.Assuming $K_{sp} = 1$ for comparison:For $M_2X$,$s = (1/4)^{1/3} \approx 0.63$.For $MX$,$s = (1)^{1/2} = 1.0$.For $MX_3$,$s = (1/27)^{1/4} \approx 0.44$.Comparing the values,the order of solubility is $MX > M_2X > MX_3$.

The solubility products $(K_{sp})$ of three sparingly soluble salts $M_2X$,$MX$,and $MX_3$ are identical. What will be the order of their solubilities $(s)$?

Similar Questions

For a sparingly soluble salt $A_pB_q$,what is the relationship between its solubility $(S)$ and solubility product $(K_{sp})$?

What is the molar solubility of $Fe(OH)_2$ $(K_{sp} = 8.0 \times 10^{-16})$ at $pH$ $13.0$ ?

An excess of $Ag_{2}CrO_{4(s)}$ is added to a $5 \times 10^{-3} \ M$ $K_{2}CrO_{4}$ solution. The concentration of $Ag^{+}$ in the solution is closest to
[ Solubility product for $Ag_{2}CrO_{4} = 1.1 \times 10^{-12}$ ]

Assertion : Addition of silver ions to a mixture of aqueous sodium chloride and sodium bromide solution will first precipitate $AgBr$ rather than $AgCl$.
Reason : $K_{sp}$ of $AgCl < K_{sp}$ of $AgBr$.

Find out the solubility of $Ni(OH)_2$ in $0.1 \ M \ NaOH$.
Given that the solubility product constant $(K_{sp})$ of $Ni(OH)_2$ is $2 \times 10^{-15}$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module