   Chapter 17, Problem 99GQ

Chapter
Section
Textbook Problem

The Ca2+ ion in hard water can be precipitated as CaCO3 by adding soda ash, Na2CO3. If the calcium ion concentration in hard water is 0.010 M and if the Na2CO3 is added until the carbonate ion concentration is 0.050 M, what percentage of the calcium ions has been removed from the water? (You may neglect carbonate ion hydrolysis.)

Interpretation Introduction

Interpretation:

Percentage of calcium ions removed from the hard water containing Ca2+ ions by adding soda ash Na2CO3 in the water has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximum amount of salt that can be dissolved in definite amount of solvent. It is expressed in moles per liter or grams per liter. Solubility in terms of moles per liter is called molar solubility and is defined as the number of moles of solute (salt) dissolved in per liter of solution.

Solubility product constant Ksp is an equilibrium constant and is defined as the product of the equilibrium concentration of the ions of the salt raised to the power of their coefficients in the balanced chemical equation.

The expression for Ksp of a salt is given as,

AxBy(s)xAy+(aq)+yBx(aq)Ksp=[Ay+]x[Bx]y

Explanation

Percentage of calcium ions removed from the hard water is calculated below.

Given:

Solubility product constant Ksp for CaCO3 is 3.4×109.

The concentration of carbonate ions added is 0.050 M.

The concentration of calcium ions present in the hard water is 0.01 M.

CaCO3 dissociates as follows in water,

CaCO3(s)Ca2+(aq)+CO32(aq)

The expression for Ksp,

Ksp=[Ca2+][CO32] (1)

When the concentration of carbonate ion in the solution is 0.050 M, the concentration of Calcium ions removed is,

[Ca2+]=Ksp[CO32]=3

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