Chapter 5, Problem 5.BE

a)

Interpretation Introduction

Interpretation:

The expression for the final concentration ${\left[\text{Ni}\right]}_{\text{f}}$ after adding $\text{25}\text{.0}\text{mL}$ of unknown to $\text{0}\text{.500}\text{mL}$ of standard has to be expressed.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

$\text{Standard}\text{addition}\text{equation=}\frac{{\left[\text{X}\right]}_{\text{i}}}{{\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}}\text{=}\frac{{\text{I}}_{\text{X}}}{{\text{I}}_{\text{S+X}}}\mathrm{...}\text{(1)}$

Where,

${\left[\text{X}\right]}_{\text{i}}$= The concentration of analyte in initial solution.

${\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}$= The concentration of analyte plus standard in final solution.

${\text{I}}_{\text{X}}$= Signal from the initial solution

${\text{I}}_{\text{S+X}}$=signal from the final solution.

For an initial volume ${\text{V}}_{\text{0}}$ of unknown and added volume ${\text{V}}_{\text{s}}$ of standard with concentration ${\left[\text{S}\right]}_{\text{i}}$,

The total volume is

${\text{V=V}}_{\text{0}}{\text{+V}}_{\text{s}}$

And the concentration of equation $\text{1}$ can be written as

$\begin{array}{l}{\left[\text{X}\right]}_{\text{f}}\text{=}{\left[\text{X}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{0}}}{\text{V}}\right)\\ {\left[\text{S}\right]}_{\text{f}}\text{=}{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{\text{V}}\right)\end{array}$

Explanation of Solution

Given,

Initial volume ( ${\text{V}}_{\text{i}}$) = $\text{25}\text{.0}\text{mL}$

Final volume ( ${\text{V}}_{\text{f}}$) =25.5mL

Final concentration ${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{f}}$= ${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{i}}}{{\text{V}}_{\text{f}}}\right)$

${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{f}}$= ${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}\left(\frac{\text{25}\text{.0}}{\text{25}\text{.5}}\right)$= $\text{0}\text{.984}{\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}$

Conclusion

The expression for the final concentration ${\left[\text{Ni}\right]}_{\text{f}}$ after adding $\text{25}\text{.0}\text{mL}$ of unknown to $\text{0}\text{.500}\text{mL}$ of standard has to be expressed.

b)

Interpretation Introduction

Interpretation:

The final concentration of added standard $\left[{\text{Ni}}^{\text{2+}}\right]$ and designated as ${\left[\text{S}\right]}_{\text{f}}$ has to be expressed.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

$\text{Standard}\text{addition}\text{equation=}\frac{{\left[\text{X}\right]}_{\text{i}}}{{\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}}\text{=}\frac{{\text{I}}_{\text{X}}}{{\text{I}}_{\text{S+X}}}\mathrm{...}\text{(1)}$

Where,

${\left[\text{X}\right]}_{\text{i}}$= The concentration of analyte in initial solution.

${\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}$= The concentration of analyte plus standard in final solution.

${\text{I}}_{\text{X}}$= Signal from the initial solution

${\text{I}}_{\text{S+X}}$=signal from the final solution.

For an initial volume ${\text{V}}_{\text{0}}$ of unknown and added volume ${\text{V}}_{\text{s}}$ of standard with concentration ${\left[\text{S}\right]}_{\text{i}}$,

The total volume is

${\text{V=V}}_{\text{0}}{\text{+V}}_{\text{s}}$

And the concentration of equation $\text{1}$ can be written as

$\begin{array}{l}{\left[\text{X}\right]}_{\text{f}}\text{=}{\left[\text{X}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{0}}}{\text{V}}\right)\\ {\left[\text{S}\right]}_{\text{f}}\text{=}{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{\text{V}}\right)\end{array}$

Explanation of Solution

Given,

${\left[\text{S}\right]}_{\text{f}}\text{=}{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{\text{V}}\right)$

Where,

$\begin{array}{l}{\text{V}}_{\text{s}}\text{=0}\text{.500}\\ \text{V=25}\text{.5}\\ {\left[\text{S}\right]}_{\text{i}}\text{=0}\text{.0287}\text{M}\end{array}$

Substituting the above values we will get

${\left[\text{S}\right]}_{\text{f}}\text{=}{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{\text{V}}\right)$

${\left[\text{S}\right]}_{\text{f}}\text{=0}\text{.0287}\text{M}\left(\frac{\text{0}\text{.500}}{\text{25}\text{.5}}\right)\text{=0}\text{.005627}\text{M}$

Conclusion

The final concentration of added standard $\left[{\text{Ni}}^{\text{2+}}\right]$ and designated as ${\left[\text{S}\right]}_{\text{f}}$ was expressed as $\text{0}\text{.005627}\text{M}$.

c)

Interpretation Introduction

Interpretation:

The concentration of ${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}$ in the unknown has to be calculated.

Concept Introduction:-

Standard addition,

Standard addition are the addition of known quantities of analyte to unknown solution, with the increase in linearity we can deduce that how much analyte present in the original unknown solution.

$\text{Standard}\text{addition}\text{equation=}\frac{{\left[\text{X}\right]}_{\text{i}}}{{\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}}\text{=}\frac{{\text{I}}_{\text{X}}}{{\text{I}}_{\text{S+X}}}\mathrm{...}\text{(1)}$

Where,

${\left[\text{X}\right]}_{\text{i}}$= The concentration of analyte in initial solution.

${\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}$= The concentration of analyte plus standard in final solution.

${\text{I}}_{\text{X}}$= Signal from the initial solution

${\text{I}}_{\text{S+X}}$=signal from the final solution.

For an initial volume ${\text{V}}_{\text{0}}$ of unknown and added volume ${\text{V}}_{\text{s}}$ of standard with concentration ${\left[\text{S}\right]}_{\text{i}}$,

The total volume is

${\text{V=V}}_{\text{0}}{\text{+V}}_{\text{s}}$

And the concentration of equation $\text{1}$ can be written as

$\begin{array}{l}{\left[\text{X}\right]}_{\text{f}}\text{=}{\left[\text{X}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{0}}}{\text{V}}\right)\\ {\left[\text{S}\right]}_{\text{f}}\text{=}{\left[\text{S}\right]}_{\text{i}}\left(\frac{{\text{V}}_{\text{S}}}{\text{V}}\right)\end{array}$

Explanation of Solution

Given,

$\text{Standard}\text{addition}\text{equation=}\frac{{\left[\text{X}\right]}_{\text{i}}}{{\left[\text{S}\right]}_{\text{f}}\text{+}{\left[\text{X}\right]}_{\text{f}}}\text{=}\frac{{\text{I}}_{\text{X}}}{{\text{I}}_{\text{S+X}}}\mathrm{...}\text{(1)}$

Where,

$\begin{array}{l}{\text{I}}_{\text{X}}\text{=2}\text{.36μΑ}\\ {\text{I}}_{\text{X+S}}\text{=3}\text{.79μΑ}\\ {\left[\text{S}\right]}_{\text{f}}\text{=0}\text{.005627M}\\ {\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{f}}\text{=0}\text{.9804}{\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}\end{array}$

$\text{Standard}\text{addition}\text{equation=}\frac{{\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}}{\text{0}\text{.005627+0}\text{.9804}{\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}}\text{=}\frac{\text{2}\text{.36}\text{μΑ}}{\text{3}\text{.79}\text{μΑ}}$

${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}\text{=9}{\text{.00×10}}^{\text{-4}}\text{M}$

Conclusion

The concentration of ${\left[{\text{Ni}}^{\text{2+}}\right]}_{\text{i}}$ in the unknown was calculated as $\text{9}{\text{.00×10}}^{\text{-4}}\text{M}$.