If spiked serum gave a signal of $\mathbf{6}\mathbf{.}\mathbf{50}\mathbf{mV}$, what was the original concentration of ${\mathbf{Na}}^{\mathbf{+}}$?

Standard addition:

Standard addition is 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

$$\begin{gathered} =\frac{{\left[\mathrm{X}\right]}_1}{{\left|\mathrm{S}\right|}_1+{\left|\mathrm{X}\right|}_1} \\ =\frac{{\mathrm{I}}_X}{{\mathrm{I}}_{\mathrm{S}+\mathrm{X}}} \end{gathered}$$

Where,

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

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

${\mathrm{I}}_{\mathrm{x}}=$Signal from the initial solution

localid="1663378201584" ${\left[\mathrm{X}\right]}_{\mathrm{i}}=$The concentration of analyte in initial solution.

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

${\mathrm{I}}_{\mathrm{x}}=$Signal from the initial solution

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

For an initial volume of unknown ${\mathrm{V}}_0$and an extra volume of standard ${\mathrm{V}}_{\mathrm{s}}$with concentration [S],

The total volume is

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

And the concentration,

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

Protein serum containing ${\mathrm{Na}}^{+}$produced a signal of $4.27\mathrm{mV}$in an atomic analysis.

$5.00\mathrm{mL}$of $2.08\mathrm{MNaCI}$were added to $95.0\mathrm{mL}$of serum.

Serum gave a signal of $6.50\mathrm{mV}$.

The final concentration of ${\mathrm{Na}}^{+}$after dilution with the standard is

$$\begin{gathered} {\left[\mathrm{X}\right]}_{\mathrm{I}}={\left[\mathrm{X}\right]}_{\mathrm{I}}\left(\frac{{\mathrm{V}}_0}{{\mathrm{V}}_{\mathrm{f}}}\right) \\ ={\left[\mathrm{X}\right]}_{\mathrm{I}}\left(\frac{95.0}{100.0}\mathrm{mL}\right) \end{gathered}$$

The final concentration of added standard is

$$\begin{gathered} {\left[\mathrm{S}\right]}_{\mathrm{f}}={\left[\mathrm{S}\right]}_{\mathrm{I}}\left(\frac{{\mathrm{V}}_{\mathrm{s}}}{\mathrm{V}}\right) \\ =\left[2.08\mathrm{M}\right]\left(\frac{5.00}{100.0}\mathrm{mL}\right) \\ =0.104\mathrm{M} \end{gathered}$$

Standard addition equation

$$\begin{gathered} =\frac{{\left[\mathrm{X}\right]}_{\mathrm{i}}}{{\left[\mathrm{S}\right]}_{\mathrm{f}}+{\left[\mathrm{X}\right]}_{\mathrm{i}}}=\frac{{\mathrm{I}}_{\mathrm{x}}}{{\mathrm{I}}_{\mathrm{s}\mathrm{x}}} \\ =\frac{{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}}{\left[0.104\mathrm{M}\right]+0.950{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}} \\ =\frac{4.27\mathrm{mV}}{6.50\mathrm{mV}} \\ =\frac{{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}}{\left[0.104\mathrm{M}\right]+0.950{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}} \\ =0.65692\mathrm{mV} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Na}}^{+}\right]=0.65692\mathrm{mV}\left(\left[0.104\mathrm{M}\right]+0.950{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}\right) \\ \left[{\mathrm{Na}}^{+}\right]=0.06832+0.6240{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}} \end{gathered}$$

$$\begin{gathered} \left[{\mathrm{Na}}^{+}\right]-0.6240{\left[{\mathrm{Na}}^{+}\right]}_{\mathrm{i}}=0.06832\left[{\mathrm{Na}}^{+}\right]\left(1-06240\right) \\ =0.06832 \\ \left[{\mathrm{Na}}^{+}\right]\left(0.3759\right)=0.06832 \\ \left[{\mathrm{Na}}^{+}\right]=\frac{0.06832}{0.3759} \\ =0.181750\mathrm{M} \\ =0.182\mathrm{M} \end{gathered}$$

Hence, the original concentration of ${\mathrm{Na}}^{+}$was calculated as $0.182\mathrm{M}$.