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

The solubility product of silver bromide $(AgBr)$ is $5.0 \times 10^{-13}$. What is the minimum amount of potassium bromide $(KBr)$ that must be added to $1 \ L$ of $0.05 \ M$ silver nitrate $(AgNO_3)$ solution to initiate the precipitation of $AgBr$? (Molar mass of $KBr = 120 \ g \ mol^{-1}$)

The molar solubility of $PbI_2$ in $0.2 \ M \ Pb(NO_3)_2$ solution in terms of $K_{sp}$ (solubility product) is

The solubility product of $Hg_2Cl_2$ in water at $25 \, ^\circ C$ is $3.2 \times 10^{-17} \, mol^3 \, dm^{-9}$. The solubility of $Hg_2Cl_2$ at $25 \, ^\circ C$ is:

For a metal hydroxide with an oxidation state of $+3$,the solubility is $1 \times 10^{-5} \ mol/L$. What is its solubility product $(K_{sp})$?

What is the $pH$ of a saturated solution of $Cu(OH)_2$? $(K_{sp} = 4.0 \times 10^{-6})$

Vedclass Test Series

Exam Paper Generator

Online Exam Module