A sodium selective electrode indicates a 34.2 mV response to a 2.50 x 10-4 M Na+ standard solution. Assuming ideal electrode response, what is the sodium concentration of a sample when the electrode indicates a 35.2 mV response?
To determine the sodium concentration of a sample when the electrode indicates a 35.2 mV response, we can use the Nernst equation, which relates the electrode potential to the concentration of the ion being measured. The Nernst equation for the sodium selective electrode is given by:
E = E° - (RT / nF) * ln([Na+])
Where:
E = electrode potential
E° = standard electrode potential (in this case, 34.2 mV)
R = ideal gas constant (8.314 J/(mol·K))
T = temperature (in Kelvin)
n = number of electrons involved in the reaction (in this case, 1 for sodium)
F = Faraday's constant (96,485 C/mol)
[Na+] = sodium concentration (in mol/L)
We need to rearrange the equation to solve for [Na+]. Rearranging gives us:
ln([Na+]) = (E° - E) * (nF / RT)
Taking the antilogarithm (exponentiation of both sides), we have:
[Na+] = e^((E° - E) * (nF / RT))
Now we can plug in the given values:
E° = 34.2 mV = 34.2 x 10^(-3) V
E = 35.2 mV = 35.2 x 10^(-3) V
R = 8.314 J/(mol·K)
T = temperature (in Kelvin - need to know the value)
n = 1
F = 96,485 C/mol
Suppose we assume a temperature of 298 K (room temperature), then we have:
[Na+] = e^((34.2 x 10^(-3) - 35.2 x 10^(-3)) * (1 * 96,485) / (8.314 * 298))
Now we can calculate the value of [Na+] using a scientific calculator or computer program.