Chapter 7, Problem 7.EE

Interpretation Introduction

Interpretation:

The ${\text{pAg}}^{\text{+}}$ values for various volumes of silver ions of given titration has to be calculated.  Also plot a graph for the titration.

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 for various volumes of silver ions of given titration.

Given,

$\begin{array}{l}\text{0}{\text{.05000M Br}}^{\text{-}}\\ \\ \text{0}{\text{.05000M Cl}}^{\text{-}}\\ \\ \text{0}{\text{.08454M AgNO}}_{\text{3}}\end{array}$

The first equivalence point where silver bromide precipitates is calculated as

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

$\begin{array}{l}{\text{V}}_{\text{e}}\text{×}\text{0}\text{.08454M}\text{=}\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0500M}\\ \\ {\text{V}}_{\text{e}}\text{=}\frac{\text{0}\text{.040}\text{L}\text{×}\text{0}\text{.0500M}}{\text{0}\text{.08454M}}\text{=}\text{0}\text{.023366}\text{L}\text{=}\text{23}\text{.66}\text{mL}\end{array}$

Silver bromide is partially precipitated upto $\text{23}\text{.66}\text{mL}$ and more bromide ions are still in solution.

When $\text{2}\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Br}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-13}}}{\left(\frac{\text{23}\text{.66}\text{mL}\text{-}\text{2}\text{.00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{42}\text{.00}\text{mL}}\right)}\\ \\ \text{=}\text{1}\text{.15}\text{×}{\text{10}}^{\text{-11}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[1}\text{.15}\text{×}{\text{10}}^{\text{-11}}\text{]}\\ \\ \text{=}\text{10}\text{.94}\end{array}$

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

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Br}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-13}}}{\left(\frac{\text{23}\text{.66}\text{mL}\text{-}10.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{50}\text{.00}\text{mL}}\right)}\\ \\ \text{=}2.188\text{×}{\text{10}}^{\text{-20}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}2.188\text{×}{\text{10}}^{\text{-20}}\text{]}\\ \\ \text{=}\text{19}\text{.66}\end{array}$

When $22\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Br}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-13}}}{\left(\frac{\text{23}\text{.66}\text{mL}\text{-}22.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{62}\text{.00}\text{mL}}\right)}\\ \\ \text{=}2.19\text{×}{\text{10}}^{\text{-10}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}2.19\text{×}{\text{10}}^{\text{-10}}\text{]}\\ \\ \text{=}\text{9}\text{.66}\end{array}$

When $23\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Br}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{5}\text{.0}\text{×}{\text{10}}^{\text{-13}}}{\left(\frac{\text{23}\text{.66}\text{mL}\text{-}23.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{63}\text{.00}\text{mL}}\right)}\\ \\ \text{=}5.623\text{×}{\text{10}}^{\text{-10}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}5.623\text{×}{\text{10}}^{\text{-10}}\text{]}\\ \\ \text{=}\text{9}\text{.25}\end{array}$

Silver chloride starts to precipitate beyond first equivalence point.

When $24\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Cl}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{1}\text{.8}\text{×}{\text{10}}^{\text{-10}}}{\left(\frac{\text{47}\text{.32}\text{mL}\text{-}24.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{64}\text{.00}\text{mL}}\right)}\\ \\ \text{=}5.8\text{×}{\text{10}}^{\text{-9}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}5.8\text{×}{\text{10}}^{\text{-9}}\text{M]}\\ \\ \text{=}8.23\end{array}$

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

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Cl}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{1}\text{.8}\text{×}{\text{10}}^{\text{-10}}}{\left(\frac{\text{47}\text{.32}\text{mL}\text{-}30.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{70}\text{.00}\text{mL}}\right)}\\ \\ \text{=}8.51\text{×}{\text{10}}^{\text{-9}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}5.8\text{×}{\text{10}}^{\text{-9}}\text{]}\\ \\ \text{=}8.07\end{array}$

When $40\text{mL}$ of silver ion is added, the concentration of silver ion is

$\begin{array}{l}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[Cl}}^{\text{-}}\text{]}}\\ \\ \text{=}\frac{\text{1}\text{.8}\text{×}{\text{10}}^{\text{-10}}}{\left(\frac{\text{47}\text{.32}\text{mL}\text{-}40.\text{00}\text{mL}}{\text{23}\text{.66}\text{mL}}\right)\left(\text{0}\text{.05000}\text{M}\right)\left(\frac{\text{40}\text{.00}\text{mL}}{\text{80}\text{.00}\text{mL}}\right)}\\ \\ \text{=}2.34\text{×}{\text{10}}^{\text{-8}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[}2.34\text{×}{\text{10}}^{\text{-8}}\text{]}\\ \\ \text{=}7.63\end{array}$

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

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

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

Volume of silver ions $=\text{(60}\text{.00- 47}\text{.32) = 2}{\text{.68 mL of Ag}}^{\text{+}}$

$\begin{array}{l}{\text{[Ag }}^{\text{+}}\text{] =}\left(\frac{\text{2}\text{.68}\text{mL}}{\text{90}\text{.00}\text{mL}}\right)\text{(0}\text{.08454}\text{M)}\text{=}\text{2}\text{.5}\text{×}{\text{10}}^{\text{-3}}\text{M}\\ \\ {\text{pAg}}^{\text{+}}\text{=}\text{-log}{\text{[Ag}}^{\text{+}}\text{]}\text{=}\text{-log[2}\text{.5}\text{×}{\text{10}}^{\text{-3}}\text{]}\\ \\ \text{=}\text{2}\text{.60}\end{array}$

The graph of ${\text{pAg}}^{\text{+}}\text{Vs}{\text{V}}_{{\text{Ag}}^{\text{+}}}$ is plotted using the above calculated ${\text{pAg}}^{\text{+}}$.

Figure 1