(C) The solubility product expression for $Mg(OH)_2$ is given by: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Given $K_{sp} = 5.6 \times 10^{-12}$ and $[Mg^{2+}] =

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

(C) The solubility product expression for $Mg(OH)_2$ is given by: $K_{sp} = [Mg^{2+}][OH^{-}]^2$ Given $K_{sp} = 5.6 \times 10^{-12}$ and $[Mg^{2+}] = 10^{-10} \ M$. Substituting the values: $5.6 \times 10^{-12} = (10^{-10}) \times [OH^{-}]^2$ $[OH^{-}]^2 = \frac{5.6 \times 10^{-12}}{10^{-10}} = 5.6 \times 10^{-2}$ $[OH^{-}] = \sqrt{5.6 \times 10^{-2}} \approx 0.2366 \ M \approx 0.24 \ M$ Thus,the concentration of $OH^{-}$ is $0.24 \ M$.

Class 11 Chemistry 6-2.Equilibrium-II (Ionic Equilibrium)Solubility product

Difficult

$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

KVPY 2017 All KVPY

A
$0.056$
B
$0.12$
C
$0.24$
D
$0.025$

Explore More

Browse All

Question Bank

All Subjects

Class 11 Questions

Class 11

Chemistry Questions

Chapter

6-2.Equilibrium-II (Ionic Equilibrium)

Subtopic

Solubility product

Previous Year

KVPY 2017 Paper All KVPY years →

If the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$, the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

DifficultAP EAMCET 2014

View Solution

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^{2} \ equivalent^{-1}$. The $K_{sp}$ of the $BaSO_4$ will be

Medium

View Solution

If the solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$ in water,its solubility in moles per litre is

Medium

View Solution

Zinc carbonate is precipitated from zinc sulphate solution by the addition of

Medium

View Solution

Sodium sulfate is soluble in water,whereas barium sulfate is sparingly soluble. Why?

Medium

View Solution

Vedclass Products

For Students

Vedclass Test Series

Mock tests in real JEE/NEET style with performance analysis. 5-day free trial.

Start Free Trial

For Teachers

Exam Paper Generator

Generate Set A/B/C/D exam papers from 7.5L+ questions in 2 minutes. 3 chapters free.

Try Free

For Institutes

Online Exam Module

Live online exams with unlimited students, 360° analytics & white-label branding.

See Demo

$Mg(OH)_2$ is precipitated when $NaOH$ is added to a solution of $Mg^{2+}$. If the final concentration of $Mg^{2+}$ is $10^{-10} \ M$,the concentration of $OH^{-} \ (M)$ in the solution is (Given: Solubility product for $Mg(OH)_2 = 5.6 \times 10^{-12}$)

Similar Questions

If the solubility product $(K_{sp})$ of $Ni(OH)_2$ is $1.9 \times 10^{-15}$, the molar solubility of $Ni(OH)_2$ in $1.0 \ M \ NaOH$ is:

The conductivity of a saturated solution of $BaSO_4$ is $3.06 \times 10^{-6} \ \Omega^{-1} \ cm^{-1}$ and its equivalent conductance is $1.53 \ \Omega^{-1} \ cm^{2} \ equivalent^{-1}$. The $K_{sp}$ of the $BaSO_4$ will be

If the solubility product of $BaSO_4$ is $1.5 \times 10^{-9}$ in water,its solubility in moles per litre is

Zinc carbonate is precipitated from zinc sulphate solution by the addition of

Sodium sulfate is soluble in water,whereas barium sulfate is sparingly soluble. Why?

Vedclass Test Series

Exam Paper Generator

Online Exam Module