K_{sp} for Cr(OH)_3 is 1.6 times 10^{-30}. Wh | vedclass

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

Question

(A) The dissociation of $Cr(OH)_3$ is represented as: $Cr(OH)_{3(s)} \rightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$Let the molar solubility be $s \

Vedclass · Accepted Answer

$\sqrt[4]{\frac{1.6 	imes 10^{-30}}{27}}$

Vedclass · Answer

(A) The dissociation of $Cr(OH)_3$ is represented as: $Cr(OH)_{3(s)} ightleftharpoons Cr^{3+}_{(aq)} + 3OH^-_{(aq)}$Let the molar solubility be $s \ mol/L$.At equilibrium,the concentration of $Cr^{3+}$ is $s$ and $OH^-$ is $3s$.The solubility product expression is: $K_{sp} = [Cr^{3+}][OH^-]^3$Substituting the values: $K_{sp} = (s)(3s)^3 = 27s^4$Given $K_{sp} = 1.6 	imes 10^{-30}$,we have: $27s^4 = 1.6 	imes 10^{-30}$Solving for $s$: $s^4 = \frac{1.6 	imes 10^{-30}}{27}$$s = \sqrt[4]{\frac{1.6 	imes 10^{-30}}{27}}$

$K_{sp}$ for $Cr(OH)_3$ is $1.6 \times 10^{-30}$. What is the molar solubility of this salt in water?

Similar Questions

In an aqueous solution containing $AgCl_{(s)}$,$AgBr_{(s)}$,and $AgI_{(s)}$,find the ratio $[Cl^{-}] : [Br^{-}] : [I^{-}]$ at equilibrium if $K_{sp}(AgCl) = 10^{-10}$,$K_{sp}(AgBr) = 10^{-13}$,and $K_{sp}(AgI) = 10^{-16}$.

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

The solubility of silver chromate is $1.992 \times 10^{-2} \ g/L$. The molar mass of $Ag_2CrO_4$ is $332 \ g/mol$. What is the molar concentration of $Ag^{+}$ in a saturated solution of $Ag_2CrO_4$?

On passing $H_2S$ gas through a solution of $Cu^{2+}$ and $Zn^{2+}$ ions,$CuS$ is precipitated first because

The solubility product of $Hg_2I_2$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module