Chapter 7, Problem 7.19P

(a)

Interpretation Introduction

Interpretation:

The ${\text{pCa}}^{\text{2+}}$ value before equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ 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,

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

Answer to Problem 7.19P

The ${\text{pCa}}^{\text{2+}}$ value before equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $6.81$.

Explanation of Solution

To calculate: The ${\text{pCa}}^{\text{2+}}$ value before equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Given,

$\begin{array}{l}\text{25}\text{.00 mL of 0}{\text{.03110 M Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\\ \\ \text{10}\text{.00}\text{mL}\text{of}\text{0}{\text{.02570M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\end{array}$

The titration reaction of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is as follows

${\text{Ca}}^{\text{2+}}{\text{ + C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{ CaC}}_{\text{2}}{\text{O}}_{\text{4}}{}_{(s)}$

Before equivalence point, more moles of oxalate ion present in solution than calcium ions.  The strength of unprecipitated oxalate ion is calculated as

$\begin{array}{l}{\text{Moles of C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}{\text{ = original moles of C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}{\text{ - moles of Ca}}^{\text{2+}}\text{ added}\\ \\ \text{=}\text{(0}\text{.025 L)(0}\text{.03110 mol/L) - (0}\text{.010 L)(0}\text{.02570 mol/L)}\\ \\ \text{=}\text{0}\text{.0005205}\text{mol}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\end{array}$

Total volume of both solution is $\text{0}\text{.035}\text{L}\text{(25}\text{.00 mL + 10}\text{.00 mL)}$

The concentration of iodide ion is

${\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\text{]}\text{=}\frac{\text{0}\text{.0005205}\text{mol}}{\text{0}\text{.035}\text{L}}\text{=}\text{0}\text{.01487}\text{M}$

The concentration of silver ion which is in equilibrium with iodide ion is

${\text{[Ca}}^{\text{2+}}\text{]}\text{=}\frac{{\text{K}}_{\text{sp}}}{{\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\text{]}}\text{=}\frac{\text{2}\text{.3}\text{×}{\text{10}}^{\text{-17}}}{\text{0}\text{.01487}}\text{=}1.5466\text{×}{\text{10}}^{\text{-7}}$

$\begin{array}{l}{\text{pCa}}^{\text{2+}}{\text{ = - log }}_{\text{10}}{\text{[Ca}}^{\text{2+}}\text{]}\\ \\ =-\log (1.5466\text{×}{\text{10}}^{\text{-7}})=6.81\end{array}$

(b)

Interpretation Introduction

Interpretation:

The ${\text{pCa}}^{\text{2+}}$ value at equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ 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,

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

Answer to Problem 7.19P

The ${\text{pCa}}^{\text{2+}}$ value at equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $4.32$.

Explanation of Solution

To calculate: The ${\text{pCa}}^{\text{2+}}$ value at equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Given,

$\begin{array}{l}\text{25}\text{.00 mL of 0}{\text{.03110 M Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\\ \\ \text{0}{\text{.02570M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\end{array}$

The titration reaction of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is as follows

${\text{Ca}}^{\text{2+}}{\text{ + C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{ CaC}}_{\text{2}}{\text{O}}_{\text{4}}{}_{(s)}$

At equivalence point, moles of oxalate present in solution is equal to calcium ions.

The concentration of silver ion and iodide ion is calculated as

$\begin{array}{l}{\text{[Ca}}^{\text{2+}}\text{]}\text{=}{\text{[ C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\text{]}\\ \\ {\text{[Ca}}^{\text{2+}}\text{]}{\text{[C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\text{]}\text{=}{\text{K}}_{\text{sp}}\\ \\ \text{(x)(x)}\text{=}2.3\text{×}{\text{10}}^{\text{-9}}\\ \\ \text{x}\text{=}\sqrt[]{2.3\text{×}{\text{10}}^{\text{-9}}}\text{=}\text{4}\text{.795}\text{×}{\text{10}}^{\text{-5}}\end{array}$

$\begin{array}{l}{\text{pCa}}^{\text{2+}}{\text{ = - log }}_{\text{10}}{\text{[Ca}}^{\text{2+}}\text{]}\\ \\ =-\log (\text{4}\text{.795}\text{×}{\text{10}}^{\text{-5}})=4.32\end{array}$

(c)

Interpretation Introduction

Interpretation:

The ${\text{pCa}}^{\text{2+}}$ value after equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ 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,

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

Answer to Problem 7.19P

The ${\text{pCa}}^{\text{2+}}$ value after equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is $2.69$

Explanation of Solution

To calculate: The ${\text{pCa}}^{\text{2+}}$ value after equivalence point for the titration of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$

Given,

$\begin{array}{l}\text{25}\text{.00 mL of 0}{\text{.03110 M Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O}}_{\text{4}}\\ \\ \text{0}{\text{.02570M Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}\end{array}$

The titration reaction of ${\text{Na}}_{\text{2}}{\text{C}}_{\text{2}}{\text{O }}_{\text{4}}\text{Vs}{\text{Ca(NO}}_{\text{3}}{\text{)}}_{\text{2}}$ is as follows

${\text{Ca}}^{\text{2+}}{\text{ + C}}_{\text{2}}{\text{O}}_{\text{4}}{}^{\text{-}}\stackrel{}{\to }{\text{ CaC}}_{\text{2}}{\text{O}}_{\text{4}}{}_{(s)}$

The volume of calcium ions at equivalence point is

$\begin{array}{l}\text{(0}\text{.025}\text{L)(0}\text{.03110}\text{M)}\text{=}{\text{V}}_{\text{e}}\text{×}\text{(0}\text{.02570)}\\ \\ {\text{V}}_{\text{e}}\text{=}\frac{\text{0}\text{.00205}}{\text{0}\text{.05110}}\text{=}30.\text{25}\text{mL}\end{array}$

Most of oxalate ion are precipitated at equivalence point.  After equivalence point, more moles of calcium ions present in solution.  The volume of calcium ion after equivalence point is $\text{35}\text{.00-}30.\text{25}\text{=}\text{4}\text{.75}\text{mL}$.

$\begin{array}{l}{\text{Moles of Ca}}^{\text{2+}}\text{ = (0}\text{.00475 L)(0}\text{.02570 mol/L)}\\ \\ \text{=}\text{0}\text{.000344}\text{mol}{\text{Ca}}^{\text{2+}}\\ \\ {\text{[Ca}}^{\text{2+}}\text{]}\text{=}\text{(0}\text{.0001221}\text{mol}\text{)/(0}\text{.060L)}\text{=}\text{0}\text{.0020346}\end{array}$

$\begin{array}{l}{\text{pCa}}^{\text{2+}}{\text{ = - log }}_{\text{10}}{\text{[Ca}}^{\text{2+}}\text{]}\\ \\ =-\log (\text{0}\text{.0020346})=2.69\end{array}$