   # What mass of Ca(NO 3 ) 2 must be added to 1.0 L of a 1.0- M HF solution to begin precipitation of CaF 2 ( s )? For CaF 2 , K sp = 4.0 × 10 −11 and K a for HF = 7.2 × 10 −4 . Assume no volume change on addition of Ca(NO 3 ) 2 ( s ). ### 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 81AE
Textbook Problem
101 views

## What mass of Ca(NO3)2 must be added to 1.0 L of a 1.0-M HF solution to begin precipitation of CaF2(s)? For CaF2, Ksp = 4.0 × 10−11 and Ka for HF = 7.2 × 10−4. Assume no volume change on addition of Ca(NO3)2(s).

Interpretation Introduction

Interpretation: The mass of Ca(NO3)2 that must be added to 1.0L solution of 1.0M HF to begin the precipitation of CaF2 is to be calculated.

Concept introduction: Solubility is defined as the maximum amount of solute that can dissolve at a certain amount of solvent at certain temperature. 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 by the formula,

Ksp=[A]x[B]y

### Explanation of Solution

Explanation

To determine: The mass of Ca(NO3)2 that must be added to 1.0L solution of 1.0M HF to begin the precipitation of CaF2 .

The equation to calculate the Ksp for CaF2 is, Ksp=[Ca2+][F]2

The reaction at equilibrium is,

CaF2(s)Ca2+(aq)+2F(aq)

Formula

The solubility product of CaF2 is calculated as,

Ksp=[Ca2+][F]2

Where,

• Ksp is solubility product.
• [Ca2+] is equilibrium concentration of Ca2+ .
• [F] is equilibrium concentration of F .

The [F] is 2.7×10-2M_ .

The dissociation reaction of HF is,

HFH++F

The ICE table is formed for the stated reaction.

HF(aq)H+(aq)+F(aq)Initialconcentration1.000Changex+x+xEquilibriumconcentration1.0xxx

The equilibrium concentration of [HF] is (1.0x)M .

The equilibrium concentration of [H+] is xM .

The equilibrium concentration of [F] is (0.020+x)M .

At equilibrium, the equilibrium constant expression is expressed by the formula,

Ka=ConcentrationofproductsConcentrationofreactants

Where,

• Ka is the acid dissociation constant.

The equilibrium constant expression for the given reaction is,

Ka=[H+][F][HF]

The given value of Ka is 7.2×104 .

Substitute the value of Ka , [HF] , [F] and [H+] in the above equation.

7.2×104=(x)(x)(1.0x)x2+(7.2×104)x7.2×104=0x=2.7×10-2M_

Therefore, the [F] is 2

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