   # Approximately 0.14 g nickel(II) hydroxide, Ni(OH) 2 ( s ), dissolves per liter of water at 20°C. Calculate K sp for Ni(OH) 2 ( s ) at this temperature. ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243

#### Solutions

Chapter
Section ### Chemistry: An Atoms First Approach

2nd Edition
Steven S. Zumdahl + 1 other
Publisher: Cengage Learning
ISBN: 9781305079243
Chapter 15, Problem 23E
Textbook Problem
318 views

## Approximately 0.14 g nickel(II) hydroxide, Ni(OH)2(s), dissolves per liter of water at 20°C. Calculate Ksp for Ni(OH)2(s) at this temperature.

Interpretation Introduction

Interpretation: The mass of Ni(OH)2 dissolves per litre of water is given. By using this value, the solubility product of Ni(OH)2 is to be calculated.

Concept introduction: The solubility product, Ksp is the equilibrium constant that is applied when salt partially dissolve in a solvent. The solubility product of dissociation of AxBy is calculated as,

Ksp=[A]x[B]y

### Explanation of Solution

Explanation

To determine: The solubility product of Ni(OH)2 from the given mass.

Solubility of Ni(OH)2 is 1.5×103mol/L_ .

Given

Mass of Ni(OH)2 is 0.14gmol/L .

Volume of solution is 1L .

Molar mass of Ni(OH)2 is 92.71g .

Formula

The solubility of Ni(OH)2 is calculated using the formula,

SolubilityofNi(OH)2=MassofNi(OH)2Volume(L)×1molNi(OH)2MolarmassofNi(OH)2

Substitute the values of mass and molar mass of Ni(OH)2 and volume of solution in the above equation.

SolubilityofNi(OH)2=MassofNi(OH)2Volume(L)×1molNi(OH)2MolarmassofNi(OH)2=0.14g1L×1molNi(OH)292.71g=1.5×103mol/L_

The concentration of Ni2+ is 1.5×103mol/L_ .

Solubility of Ni(OH)2 is 1.5×103mol/L .

Since, solid Ni(OH)2 is placed in contact with water. Therefore, compound present before the reaction is Ni(OH)2 and H2O . The dissociation reaction of Ni(OH)2 is,

Ni(OH)2Ni2+(aq)+2OH(aq)

Since, Ni(OH)2 does not dissolved initially, hence,

[Ni2+]initial=[OH]initial=0

The concentration at equilibrium can be calculated from the measured solubility of Ni(OH)2 . If 1.5×103mol/L of Ni(OH)2 is dissolved in 1.0L of solution, the change in solubility will be equal to 1.5×103mol/L . The reaction is,

Ni(OH)2(s)Ni2+(aq)+2OH(aq)

Therefore,

1.5×103mol/LNi(OH)2(1.5×103mol/L)Ni2++2(1

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