Chapter 17.3, Problem 22PP

(a)

Interpretation Introduction

Interpretation:

When solubility product is given then the concentration $\text{Ag}$ ion is to be calculated.

Concept introduction:

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

Answer to Problem 22PP

The concentration of the ion $\text{Ag}$ 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

$\text{AgBr}\rightleftharpoons {\text{Ag}}^{\text{1+}}{\text{+ Br}}^{\text{1-}}$

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

$\text{Ksp = }\left[{\text{Ag}}^{\text{1+}}\right]\text{. }\left[{\text{Br}}^{\text{1-}}\right]$

$\text{= x × x}$

${\text{ = x}}^{\text{2}}$

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

Or, $\text{x = }\sqrt{\text{54}}\text{×}\sqrt{{\text{10}}^{\text{-14}}}$

= $\text{7}{\text{.348 × 10}}^{\text{-7}}$ mol/L

(b)

Interpretation Introduction

When solubility product is given then the concentration of fluorine ion is to be calculated

Interpretation:

When solubility product is given then the concentration of fluorine ion is to be calculated

Concept introduction:

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

Answer to Problem 22PP

The solubility of the fluorine ion is $\text{2}{\text{.136 × 10}}^{\text{-4}}\text{mol/L}\text{.}$

Explanation of Solution

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

${\text{CaF}}_{\text{2}}\underset{}{\rightleftharpoons }{\text{Ca}}^{\text{2+}}{\text{+ 2F}}^{\text{1-}}$

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

C $\text{= +x +2x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= x 2x }$ (equilibrium concentration)

The solubility product is given as

$\text{Ksp = }\left[{\text{Ca}}^{\text{+2}}\right]\times {\left[{\text{F}}^{\text{1-}}\right]}^{\text{2}}$

${\text{ = x × 4x}}^{\text{2}}$

${\text{= 4 x}}^{\text{3}}$

${}^{\text{3}}\sqrt{\left(\text{ Ksp / 4}\right)}\text{= x}$

Or, $\text{x = }{\left(\text{ 3}{\text{.9×10}}^{\text{-11}}\text{/4}\right)}^{\text{1/3}}$

$\text{= 2}{\text{.136×10}}^{\text{-4}}$ mol/L

(c)

Interpretation Introduction

Interpretation:

When solubility product is given, the concentration of silver ion is to be calculated

Concept introduction:

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

Answer to Problem 22PP

The concentration of silver ion is $\text{0}{\text{.65 × 10}}^{\text{-4}}$ mol/L.

Explanation of Solution

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

${\text{Ag2CrO}}_{\text{4}}\rightleftharpoons {\text{2Ag}}^{\text{1+}}{\text{+CrO}}_{\text{4}}{}^{\text{-2}}$

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

C $\text{= +2x +x }$ (change in concentration or change in amount from initial state to equilibrium)

E $\text{= 2x x }$ (equilibrium concentration)

The solubility product is given as

$\text{Ksp = }{\left[{\text{Ag}}^{\text{1+}}\right]}^{\text{2}}\text{× }\left[{\text{CrO}}_{\text{4}}{}^{\text{2-}}\right]$

${\text{= 4x}}^{\text{2}}\text{× x}$

${\text{ = 4x}}^{\text{3}}$

${}^{\text{3}}\sqrt{\text{Ksp/4}}\text{ = x}$

Or, ${\text{x = }}^{\text{3}}\sqrt{\left(\text{1}{\text{.1× 10}}^{\text{-12}}\text{/4}\right)}$

= $\text{0}{\text{.6503 × 10}}^{\text{-4}}$ mol/L