Chapter 7, Problem 7.25P

(a)

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration has to be calculated.

Concept introduction:

$\begin{array}{l}{\text{pX = - log }}_{\text{10}}\text{[X]}\\ \\ \text{p}\text{function}\\ \\ \text{[X]}\text{is}\text{concentration}\text{of}\text{X}\end{array}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Explanation of Solution

To calculate: The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}\text{.0502M KI}\\ \\ \text{0}\text{.0500M KCl}\\ \\ \text{0}{\text{.0845M AgNO}}_{\text{3}}\end{array}$

The first equivalence point where silver iodide precipitates first because ${\text{K}}_{\text{sp}}{\text{(AgI)<< K}}_{\text{sp}}\text{(AgCl)}$ is calculated as

Moles of silver ion is equal to moles of iodide ions.

$\begin{array}{l}{\text{V}}_{\text{e}}\text{×}\text{0}\text{.0845M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}\\ \\ {\text{V}}_{\text{e}}\text{=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}}{\text{0}\text{.0845M}}\text{=}\text{0}\text{.02376}\text{L}\text{=}\text{23}\text{.76}\text{mL}\end{array}$

Silver iodide is partially precipitated upto $\text{23}\text{.76}\text{mL}$ and more iodide ions are still in solution.

When $10\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[I}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{8.3\text{×}{\text{10}}^{\text{-17}}}{\left(\frac{\text{23}\text{.76}\text{mL}\text{-}10.\text{00}\text{mL}}{\text{23}\text{.76}\text{mL}}\right)\left(\text{0}\text{.0502}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{50}\text{.00}\text{mL}}\right)}\\ \\ \text{=}3.562\text{×}{\text{10}}^{\text{-15}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}3.562\text{×}{\text{10}}^{\text{-15}}\text{]}\\ \\ \text{=}14.45\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration has to be calculated.

Concept introduction:

$\begin{array}{l}{\text{pX = - log }}_{\text{10}}\text{[X]}\\ \\ \text{p}\text{function}\\ \\ \text{[X]}\text{is}\text{concentration}\text{of}\text{X}\end{array}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Explanation of Solution

To calculate: The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}\text{.0502M KI}\\ \\ \text{0}\text{.0500M KCl}\\ \\ \text{0}{\text{.0845M AgNO}}_{\text{3}}\end{array}$

The first equivalence point where silver iodide precipitates first because ${\text{K}}_{\text{sp}}{\text{(AgI)<< K}}_{\text{sp}}\text{(AgCl)}$ is calculated as

Moles of silver ion is equal to moles of iodide ions.

$\begin{array}{l}{\text{V}}_{\text{e}}\text{×}\text{0}\text{.0845M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}\\ \\ {\text{V}}_{\text{e}}\text{=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}}{\text{0}\text{.0845M}}\text{=}\text{0}\text{.02376}\text{L}\text{=}\text{23}\text{.76}\text{mL}\end{array}$

When $20\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[I}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{8.3\text{×}{\text{10}}^{\text{-17}}}{\left(\frac{\text{23}\text{.76}\text{mL}\text{-}20.\text{00}\text{mL}}{\text{23}\text{.76}\text{mL}}\right)\left(\text{0}\text{.0502}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{60}\text{.00}\text{mL}}\right)}\\ \\ \text{=}1.585\text{×}{\text{10}}^{\text{-14}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}1.585\text{×}{\text{10}}^{\text{-14}}\text{]}\\ \\ \text{=}\text{13}\text{.80}\end{array}$

(c)

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration has to be calculated.

Concept introduction:

$\begin{array}{l}{\text{pX = - log }}_{\text{10}}\text{[X]}\\ \\ \text{p}\text{function}\\ \\ \text{[X]}\text{is}\text{concentration}\text{of}\text{X}\end{array}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Explanation of Solution

To calculate: The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}\text{.0502M KI}\\ \\ \text{0}\text{.0500M KCl}\\ \\ \text{0}{\text{.0845M AgNO}}_{\text{3}}\end{array}$

The second equivalence point where silver chloride starts to precipitate is calculated as

Moles of silver ion is equal to moles of chloride ions.

$\begin{array}{l}{\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL}\text{×}\text{0}\text{.0845M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}\\ \\ {\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}}{\text{0}\text{.0845M}}\text{=}\text{0}\text{.02376}\text{L}\text{=}\text{23}\text{.76}\text{mL}\\ \\ {\text{V}}_{\text{e}}\text{=}\text{23}\text{.76}\text{mL}\text{+}\text{23}\text{.76}\text{mL}\text{=}\text{47}\text{.52}\text{mL}\end{array}$

Silver chloride starts to precipitate beyond first equivalence point.

When $30\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Cl}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{1}\text{.8}\text{×}{\text{10}}^{\text{-10}}}{\left(\frac{\text{47}\text{.52}\text{mL}\text{-}30.\text{00}\text{mL}}{\text{23}\text{.76}\text{mL}}\right)\left(\text{0}\text{.0502}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{70}\text{.00}\text{mL}}\right)}\\ \\ \text{=}8.5\text{×}{\text{10}}^{\text{-9}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}5.8\text{×}{\text{10}}^{\text{-9}}\text{M]}\\ \\ \text{=}8.07\end{array}$

(d)

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration has to be calculated.

Concept introduction:

$\begin{array}{l}{\text{pX = - log }}_{\text{10}}\text{[X]}\\ \\ \text{p}\text{function}\\ \\ \text{[X]}\text{is}\text{concentration}\text{of}\text{X}\end{array}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Explanation of Solution

To calculate: The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}\text{.0502M KI}\\ \\ \text{0}\text{.0500M KCl}\\ \\ \text{0}{\text{.0845M AgNO}}_{\text{3}}\end{array}$

The second equivalence point where silver chloride starts to precipitate is calculated as

Moles of silver ion is equal to moles of chloride ions.

$\begin{array}{l}{\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL}\text{×}\text{0}\text{.0845M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}\\ \\ {\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}}{\text{0}\text{.0845M}}\text{=}\text{0}\text{.02376}\text{L}\text{=}\text{23}\text{.76}\text{mL}\\ \\ {\text{V}}_{\text{e}}\text{=}\text{23}\text{.76}\text{mL}\text{+}\text{23}\text{.76}\text{mL}\text{=}\text{47}\text{.52}\text{mL}\end{array}$

The concentration of silver ion and chloride ion are equal at equivalence point.

$\begin{array}{l}{\text{[Ag}}^{\text{+}}{\text{][Cl}}^{\text{-}}{\text{] = x}}^{\text{2}}\text{=}{\text{K}}_{\text{sp}}\text{(for AgCl)}\\ \\ {\text{[Ag}}^{\text{+}}{\text{][Cl}}^{\text{-}}{\text{] = x}}^{\text{2}}\text{=}\text{1}\text{.8}\text{×}{\text{10}}^{\text{-10}}\\ \\ {\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{1}\text{.34}\text{×}{\text{10}}^{\text{-5}}\\ \\ {\text{pAg}}^{\text{+}}\text{ = 4}\text{.87}\end{array}$

(e)

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration has to be calculated.

Concept introduction:

$\begin{array}{l}{\text{pX = - log }}_{\text{10}}\text{[X]}\\ \\ \text{p}\text{function}\\ \\ \text{[X]}\text{is}\text{concentration}\text{of}\text{X}\end{array}$

Molarity can be defined as the moles of solute (in grams) to the volume of the solution (in litres). The molarity of a solution can be given by the formula,

$\text{Molarity(M)=}\frac{\text{Moles}\text{of}\text{solute(in}\text{g)}}{\text{Volume}\text{of}\text{solution(in}\text{L)}}$

Explanation of Solution

To calculate: The ${\text{pAg}}^{\text{+}}$ values of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}\text{.0502M KI}\\ \\ \text{0}\text{.0500M KCl}\\ \\ \text{0}{\text{.0845M AgNO}}_{\text{3}}\end{array}$

The second equivalence point where silver chloride starts to precipitate is calculated as

Moles of silver ion is equal to moles of chloride ions.

$\begin{array}{l}{\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL}\text{×}\text{0}\text{.0845M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}\\ \\ {\text{V}}_{\text{e}}\text{-}\text{23}\text{.76}\text{mL=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0502M}}{\text{0}\text{.0845M}}\text{=}\text{0}\text{.02376}\text{L}\text{=}\text{23}\text{.76}\text{mL}\\ \\ {\text{V}}_{\text{e}}\text{=}\text{23}\text{.76}\text{mL}\text{+}\text{23}\text{.76}\text{mL}\text{=}\text{47}\text{.52}\text{mL}\end{array}$

When $50\text{mL}$ of silver ion is added, there is only silver ions present in excess.

Volume of silver ions $=\text{(60}\text{.00- 47}\text{.52) = 12}{\text{.48 mL of Ag}}^{\text{+}}$

$\begin{array}{l}{\text{[Ag }}^{\text{+}}\text{] =}\left(\frac{\text{12}\text{.68}\text{mL}}{\text{90}\text{.00}\text{mL}}\right)\text{(0}\text{.0845}\text{M)}\text{=}\text{2}\text{.45}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[2}\text{.45}\text{×}{\text{10}}^{\text{-3}}\text{]}\\ \\ \text{=}\text{2}\text{.61}\end{array}$