   Chapter 17.4, Problem 17.9CYU

Chapter
Section
Textbook Problem

Calculate the solubility of AgCN in moles per liter and grams per liter. Ksp for AgCN is 6.0 × 10−17.

Interpretation Introduction

Interpretation:

Solubility of salt AgCN has to be calculated in moll1 and gl1 units using given Ksp value.

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

The 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(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

Solubility of salt AgCN in moll1 and gl1 units is calculated below.

Given:

The balanced chemical reaction  involved  is,

AgCN(s) Ag+1(aq)+ CN1(aq)

The ICE table is as follows,

EquationAgCNAg+1+CN1Initial (M)00Change (M)+s+sEquilibrium (M) ss

The value of solubility product Ksp of AgCN is 6.0×1017,

Ksp=[Ag+1][CN1]

Substitute s for [Ca2+] and for [CN1].

Ksp=s×s

Here,

• Ksp is solubility product constant
• s is solubility

Ksp=s2

Rearrange the above expression.

s=Ksp2

Again, substitute 6

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