Chapter 17.3, Problem 20PP

(a)

Interpretation Introduction

Interpretation:

When solubility product is given then the solubility in mol/L is to be calculated.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

Answer to Problem 20PP

The solubility of the compound ${\text{PbCrO}}_{\text{4}}$ mol/L is $\text{4}{\text{.795 × 10}}^{\text{-7}}$ mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

${\text{PbCrO}}_{\text{4}}\underset{}{\rightleftharpoons }{\text{Pb}}^{\text{2+}}{\text{+ CrO}}_{\text{4}}{}^{\text{2-}}$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= x x }$ (equilibrium concentration)

The solubility product is given as

$\begin{array}{l}\text{ Ksp = }\left[{\text{Pb}}^{\text{2+}}\right]\text{. }\left[{\text{CrO}}_{\text{4}}{}^{\text{2-}}\right]\hfill \\ \text{ = x }\text{. x}\hfill \\ {\text{ = x}}^{\text{2}}\hfill \\ \sqrt{\text{Ksp}}{\text{ = x}}^{}\hfill \end{array}$

Or, $\text{x = }\sqrt{\text{23}}{\text{ × 10}}^{\text{-14}}$

$\text{= 4}{\text{.795 × 10}}^{\text{-7}}$ mol/L

(b)

Interpretation Introduction

When solubility product is given then the solubility in mol/L is to be calculated

Interpretation:

The positron should be defined and described.

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

Answer to Problem 20PP

The solubility of the compound $\text{AgCl}$ in mol/L is $\text{1}{\text{.342 × 10}}^{\text{-5}}$ mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

$\text{ AgCl}\rightleftharpoons {\text{Ag}}^{\text{1+}}{\text{+ Cl}}^{\text{1-}}$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= x x }$ (equilibrium concentration)

The solubility product is given as

$\begin{array}{l}\hfill \\ \text{ Ksp = }\left[{\text{Ag}}^{\text{+}}\right]\text{. }\left[{\text{Cl}}^{\text{1-}}\right]\hfill \\ \text{ = x }\text{. x}\hfill \\ {\text{ = x}}^{\text{2}}\hfill \end{array}$

$\sqrt{\text{Ksp}}\text{ = x}$

Or, $\text{x = }\sqrt{\text{1}\text{.8}}{\text{× 10}}^{\text{-10}}$

$\text{= 1}{\text{.342 × 10}}^{\text{-5}}$ mol/L

(c)

Interpretation Introduction

Interpretation:

When solubility product is given then the solubility in mol/L is to be calculated

Concept introduction:

Solubility is the concentration of the solute dissolved in a given solvent.

Answer to Problem 20PP

The solubility of the compound ${\text{CaCO}}_{\text{3}}$ in mol/L is $\text{5}{\text{.831 × 10}}^{\text{-5}}$ mol/L.

Explanation of Solution

Solubility is the concentration of solute which dissolved in a given solvent. So, the given reaction is

${\text{CaCO}}_{\text{3}}\rightleftharpoons {\text{Ca}}^{\text{2+}}{\text{+CO}}_{\text{3}}{}^{\text{-2}}$

I $\text{= 0 0 }$ (initial concentration)

C $\text{= +x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= x x }$ (equilibrium concentration)

The solubility product is given as

$\begin{array}{l}\text{ Ksp = }\left[{\text{Ca}}^{\text{2+}}\right]\text{. }\left[{\text{CO}}_{\text{3}}{}^{\text{2-}}\right]\hfill \\ \text{ = x }\text{. x}\hfill \\ {\text{ = x}}^{\text{2}}\hfill \end{array}$

$\sqrt{\text{Ksp}}\text{ = x}$

Or, $\text{x = }\sqrt{\text{3}\text{.4}}{\text{× 10}}^{\text{-9}}$

${}^{}\text{ = 5}{\text{.831 × 10}}^{\text{-5}}$ mol/L