Standard addition graph. Students performed an experiment like that in Figure 5-7 in which each flask contained $\mathbf{25}\mathbf{.}\mathbf{00}\mathbf{mL}$of serum, varying additions of $\mathbf{2}\mathbf{.}\mathbf{640}\mathbf{}\mathbf{M}\mathbf{}\mathbf{NaCI}$standard, and a total volume of $\mathbf{50}\mathbf{.}\mathbf{00}\mathbf{mL}\mathbf{.}$

(a) Prepare a standard addition graph and find $\left[{\mathrm{Na}}^{+}\right]$in the serum.

(b) Find the standard deviation and $\mathbf{95}\mathbf{\%}$confidence interval for $\mathbf{\left[}{\mathbf{Na}}^{\mathbf{+}}\mathbf{\right]}$.

(a)

The concentration of added $\mathrm{NaCI}\mathrm{and}{\mathrm{l}}_{\mathrm{s}+\mathrm{x}}$are plotted on a spreadsheet with a axis.

Every solution is created with the same volume.

Then ${\mathrm{l}}_{\mathrm{s}+\mathrm{x}}\mathrm{vs}{\left[\mathrm{S}\right]}_{\mathrm{f}}$is plotted as ${\left[\mathrm{X}\right]}_{\mathrm{f}}=0.698\mathrm{M}$has a negative intercept.

The dilution factor $2\left(50.00\mathrm{mL}/25.0\mathrm{mL}\right)$makes the starting concentration of $\mathrm{NaCl}$bigger. In serum, the initial concentration of $\mathrm{NaCl}$was $2.000\times 0.0698\mathrm{M}=0.01396\mathrm{M}.$

The concentration of was calculated to be$0.01396\mathrm{M}.$

The concentration of added $\mathrm{NaCl}\mathrm{and}{\mathrm{l}}_{\mathrm{s}+\mathrm{x}}$are plotted on a spreadsheet with a $\mathrm{x}-\mathrm{axis}.$

The $\mathrm{x}-$intercept is determined in cell B20, and its uncertainty is estimated in cell B27, as shown in Figure 1.

$\left(0.00235\right)/\left(0.0698\right)=3.37\%$percent relative uncertainty

Because the uncertainty in the original concentration of ${\mathrm{Na}}^{+}$is significantly bigger than the relative uncertainties in volume measurement, the uncertainty in the original concentration of ${\mathrm{Na}}^{+}$is estimated to be $3.37\%.$

$\left[{\mathrm{Na}}^{+}\right]$has a sensible expression in the original serum is$0.140\left(\pm 3.37\%\right)\mathrm{M}=0.140\left(\pm 0.0047\mathrm{M}\right)$

The interval in $95\%$confidence is $\mathrm{l}=\pm \mathrm{ts}=\pm \left(3.182\right)\left(0.0047\mathrm{M}\right)=\pm 0.015\mathrm{M}$

Here, $\mathrm{t}$is for $5-2=3\mathrm{degree}$of freedom.

$0.140\left(\pm 3.37\%\right)$is the standard addition and$95\%$ confidence interval of ${\mathrm{Na}}^{+}\mathrm{is}\pm 0.015\mathrm{M}.$