91Ó°ÊÓ

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.

Standard addition equation $=\frac{{\left[\mathrm{X}\right]}_{\mathrm{i}}}{{\left[\mathrm{S}\right]}_{\mathrm{f}}{\left[\mathrm{X}\right]}_{\mathrm{f}}}=\frac{{\mathrm{l}}_{\mathrm{x}}}{{\mathrm{l}}_{\mathrm{s}+\mathrm{x}}}$

Where,

role="math" localid="1654763027013" ${\left[\mathrm{X}\right]}_{\mathrm{i}}=$The concentration of analyte in initial solution.

role="math" localid="1654763033694" ${\left[\mathrm{S}\right]}_{\mathrm{f}}+{\left[\mathrm{X}\right]}_{\mathrm{f}}=$The concentration of analyte plus standard in final solution.

role="math" localid="1654763101592" $\mathrm{l}x=$Signal from the initial solution

${\mathrm{l}}_{\mathrm{s}+\mathrm{x}}=$signal from the final solution.

For an initial volume $V_0$of unknown and added volume $V_s$of standard with concentration ${\left[\mathrm{S}\right]}_{\mathrm{i}}$,

The total volume is

$\mathrm{V}={\mathrm{V}}_0+{\mathrm{V}}_{\mathrm{s}}$

And the concentration of equation 1 can be written as

$$\begin{gathered} {\left[\mathrm{X}\right]}_{\mathrm{f}}={\left[\mathrm{X}\right]}_{\mathrm{i}}\left(\frac{{\mathrm{V}}_0}{\mathrm{V}}\right) \\ {\left[\mathrm{S}\right]}_{\mathrm{f}}={\left[\mathrm{S}\right]}_{\mathrm{i}}\left(\frac{{\mathrm{V}}_{\mathrm{s}}}{\mathrm{V}}\right) \end{gathered}$$

Initial volume$\left({\mathrm{V}}_{\mathrm{i}}\right)=25.0\mathrm{mL}$

Final volume$\left({\mathrm{V}}_{\mathrm{f}}\right)=25.5\mathrm{mL}$

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

${\left[{\mathrm{Ni}}^{2+}\right]}_{\mathrm{f}}=\left[{\mathrm{Ni}}^{2+}\right]\mathrm{i}\left(\frac{25.0}{25.0}\right)=0984\left[{\mathrm{Ni}}^{2+}\right]\mathrm{i}$

The expression for the final concentration$\left[\mathrm{Ni}\right]$after adding$25.0\mathrm{mL}$of unknown to$0.500\mathrm{mL}$ of standard has to be expressed.

b)

Given,

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

Where,

$$\begin{gathered} {\mathrm{V}}_{\mathrm{s}}=0.500 \\ \mathrm{V}=25.5 \\ {\left[\mathrm{S}\right]}_{\mathrm{i}}=0.0287\mathrm{M} \end{gathered}$$

Substituting the above values we will get,

$$\begin{gathered} {\left[\mathrm{S}\right]}_{\mathrm{f}}={\left[\mathrm{S}\right]}_{\mathrm{i}}\left(\frac{{\mathrm{V}}_{\mathrm{s}}}{\mathrm{V}}\right) \\ {\left[\mathrm{S}\right]}_{\mathrm{f}}=0.0287\mathrm{M}\left(\frac{0.500}{25.5}\right) \\ =0.005627\mathrm{M} \end{gathered}$$

c)

Standard addition equation$=\frac{{\left[\mathrm{X}\right]}_{\mathrm{i}}}{{\left[\mathrm{S}\right]}_{\mathrm{f}}+{\left[\mathrm{X}\right]}_{\mathrm{f}}}=\frac{{\mathrm{l}}_{\mathrm{x}}}{{\mathrm{l}}_{\mathrm{s}+\mathrm{x}}}$

Where,

$$\begin{gathered} {\mathrm{i}}_{\mathrm{x}}=2.36\mathrm{μ´¡} \\ {\mathrm{l}}_{\mathrm{x}+\mathrm{s}}=3.79\mathrm{μ´¡} \\ {\left[\mathrm{S}\right]}_{\mathrm{f}}=0.005627\mathrm{M} \\ {\left[{\mathrm{Ni}}^{2+}\right]}_{\mathrm{f}}=0.9804\left[{\mathrm{Ni}}^{2+}\right] \end{gathered}$$

Standard addition equation $=\frac{{\left[{\mathrm{Ni}}^{2+}\right]}_{\mathrm{i}}}{0.005627+0.9804{\left[{\mathrm{Ni}}^{2+}\right]}_{\mathrm{i}}}=\frac{2.36\mathrm{μ´¡}}{3.79\mathrm{μ´¡}}{\left[{\mathrm{Ni}}^{2+}\right]}_{\mathrm{i}}=9.00×{10}^{-4}\mathrm{M}$