A water sample was analysed using a sodium selective electrode. The electrode response was 75.0 mV. A 10.00 mL aliquot of standard 0.50 mmol/L NaCl solution was added to 10.00 mL of water sample. The well-mixed solution gave a response of 98.0 mV. What is the concentration of Na+ in the water sample?

Question

A water sample was analysed using a sodium selective electrode. The electrode response was 75.0 mV. A 10.00 mL aliquot of standard 0.50 mmol/L NaCl solution was added to 10.00 mL of water sample. The well-mixed solution gave a response of 98.0 mV. What is the concentration of Na+ in the water sample?

Answer 1

To determine the concentration of Na+ in the water sample, we can use the Nernst equation, which relates the measured electrode potential to the concentration of the analyte in solution.

The Nernst equation is given by:

E = E° + (RT/nF) * ln(C)

Where:
E is the measured electrode potential,
E° is the standard electrode potential,
R is the gas constant (8.314 J/(mol·K)),
T is the temperature in Kelvin,
n is the number of moles of electrons transferred in the redox reaction,
F is the Faraday constant (96485 C/mol),
C is the concentration of the analyte in solution.

First, we need to determine the standard electrode potential (E°). In this case, the standard electrode potential for the sodium selective electrode is not given. You need to find the specific value for the electrode being used using reference materials or consult the electrode manufacturer.

Next, we plug the values into the equation. The temperature (T) should be specified, n is usually 1 for single-electron transfer reactions, and F is a constant.

For the initial measurement, we have:
E1 = 75.0 mV

For the measurement after adding the NaCl solution, we have:
E2 = 98.0 mV
C2 = 0.5 mmol/L

Substituting these values into the Nernst equation, we have:

E1 = E° + (RT/nF) * ln(C1)

E2 = E° + (RT/nF) * ln(C1 + C2)

Subtracting the two equations, we get:

E2 - E1 = (RT/nF) * ln(C1 + C2) - (RT/nF) * ln(C1)

Simplifying further:

E2 - E1 = (RT/nF) * ln((C1 + C2)/C1)

Now, we can solve for C1, the concentration of Na+ in the water sample.

(C1 + C2)/C1 = exp((E2 - E1) * (nF/RT))

C1 = C2 / (exp((E2 - E1) * (nF/RT)) - 1)

Substituting the known values:
C2 = 0.5 mmol/L,
E1 = 75.0 mV,
E2 = 98.0 mV,
n = 1 (since it's a single-electron transfer for Na+),
F = 96485 C/mol (Faraday constant),
R = 8.314 J/(mol·K),

We can then calculate C1.