Question

A sodium ion electrode was calibrated by measuring the cell potentials of several standard Na⁺ solutions. The other electrode was the SCE. The calibration data are shown in the following table.

Log[Na+]	Ecell (mV)
-1	-165.6
0	-110.0
1	-51.8
2	6.5
3	64.5

a. Plot a calibration curve for these data.

b. Determine the slope and intercept for the regression line of this plot. Is the response of this sodium electrode Nernstian? Explain.

c. A 10.00mL aliquot of table wine was diluted to 100.0mL in a volumetric flask. A 5.00mL aliquot of this solution was then diluted to 250.0mL. This final solution gave a cell potential reading of -125.6 mV. Calculate the concentration of Na⁺ in the wine in ppm.

d. What is the answer to part (c) expressed as the molarity of sodium ion?

Answer 1

a. Calibration curve

To plot the calibration curve of this potentiometrical measurement, we will express the cell potential (Ecell) as a function of Na+ concentration (Log [Na+]) in the following table

Log [Na+]	Ecell (mV)
-1	-165.6
0	-110.0
1	-51.8
2	6.5
3	64.5

And the plot of these data is the following

Ecell (mv) Log [Na] -2 -1 - 1 2

b. Regression line

The slope and the intercept of the regression line of this plot are calculated from graphic methods such as

Ecell (mv) Log [Nat] -2 -1 1 / 2 3 Ecell= 57,67Log [Nat] - 108,95 -150 -200

Where the slope is:

And the intercept is:

b= -108.95

In direct potentiometric measurements, the cell potential is related to the analite concentration according to the Nernstian equation:

0.0592 Ecell = K -

Where K is a constant due to the potential of the indicator electrode and the junction potential, n is the number of electrons transferred and pNa is given by:

$pNa=-Log(\left [ Na^{+} \right ])$

Substituting we have:

$E_{cell}=K+\frac{0.0592}{n}Log\left ( \left [ Na^{+} \right ] \right )$

Then, we can assume that the response of this sodium electrode is Nernstian.

c. Concentration of Na+ in the wine in ppm (mg/L)

By using the regression line, we can calculate directly the Na+ concentration in the final sample measured as follows

$E_{cell}=57.67Log\left ( \left [ Na^{+} \right ] \right )-108.95$

Solving for Na+ concentration ([Na+])

$\frac{E_{cell}+108.95}{57.67}=Log\left ( \left [ Na^{+} \right ] \right )$

According to logarithmic properties

Eccl +108.95 10:57.67 = Na

Substituting data given

$10^{\frac{-125.6+108.95}{57.67}}=\left [ Na^{+} \right ]$

The molar concentration of Na+ in this sample is

$0.514M=\left [ Na^{+} \right ]$

But this final sample was diluted twice. In the second dilution we apply the dilution equation:

$\left [ Na^{+} \right ]_{concentrated} =\left [ Na^{+} \right ]_{diluted}*\frac{V_{diluted}}{V_{concentrated}}$

Where Vdiluted is 250.0mL and the Vconcentrated is 5.00mL, then:

$\left [ Na^{+} \right ]_{concentrated} =0.514M*\frac{250.0mL}{5.00mL}=25.7M$

In the first dilution we apply the same dilution equation, in the wine, the Na+ concentration is

$\left [ Na^{+} \right ]_{wine} =\left [ Na^{+} \right ]_{diluted}*\frac{V_{diluted}}{V_{wine}}$

In this case, Vdiluted is 100.0mL and the Vwine is 10.00mL

$\left [ Na^{+} \right ]_{wine} =25.7M*\frac{100.0mL}{10.00mL}=257M$

NOTICE THAT it is the answer of the part d. The molarity of soium ion in the wine is 257M

To convert it into ppm units, we will use a conversion factor

$ppm Na^{+} =\frac{257mol Na^{+}}{1L}*\frac{22.9898 g Na^{+}}{1mol\, Na^{+}}*\frac{1000mg}{1g}$

In the wine, we have:

This is a large quantity but is in the range of the potentiometric measurement. As you can see, the Log(Na+) = 3 shows that the molarity of that standard solution is 1000M