Chapter 17, Problem 44PS

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

Chapter
Section

### Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

# At 20 °C, a saturated aqueous solution of silver acetate, AgCH3CO2, contains 1.0 g of the silver compound dissolved in 100.0 mL of solution. Calculate Ksp for silver acetate. A g C H 3 C O 2 ( s ) ⇄ A g + ( a q ) + C H 3 C O 2 − ( a q )

Interpretation Introduction

Interpretation:

Solubility product constant Ksp for silver acetate, AgCH3CO2 has to be calculated.

Concept introduction:

The solubility of a salt is defined as the maximium 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.

Expression for Ksp of a salt is given as,

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

Relation between Ksp and s is derived as follows,

EquationAxByxAy++yBxInitial(M)00Change(M)+xs+ysEquilibrium(M)xsys

Ksp=[Ay+]x[Bx]yKsp=(xs)x(ys)y=xxyy(s)x+y

Rearrange the expression for s.

(s)x+y=Kspxxyy=(Kspxxyy)1/(x+y)

Here,

• x is the coefficient of cation A+y
• y is the coefficient of anion Bx
• s is the molar solubility

Ksp is calculated by using molar solubility of the salt.

Explanation

The solubility product constant Ksp for AgCH3CO2 is calculated below.

Given:

1 gram of silver acetate is dissolved in 100 ml of water.

AgCH3CO2 when dissolved in water dissociates as follows:

AgCH3CO2(s)â‡ŒAg+1(aq)+CH3COOâˆ’(aq)

Number of moles of silver acetate dissolved in 100 ml of solution or molar solubility is,

MolarÂ solubility,Â sÂ =Â (wM)V(l)

Here,

• w is the given mass of AgCH3CO2
• M is the gram molecular mass of AgCH3CO2
• V is the volume of solution in liter

MolarÂ solubility,Â sÂ =Â (1166.9)(100Â ml)=Â (1166.9)(100Â ml)(1Â l1000Â ml)=0

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