2
$\begingroup$

I know how to use the standard addition equation for a single-point analysis from an equation I got from the Analytical Chemistry book "Quantitative Chemical Analysis" 9ed by D. Harris.

However, solving for a potentiometry problem, the equation doesn't seem to be effective anymore. I keep on getting negative value for the initial concentration of the unknown sample.

The problem goes like this:

The Na+ concentration of a solution was determined by measurement with a sodium ion-selective electrode. The electrode system developed a potential of -0.2462 V when immersed in 10.0 mL of the solution of unknown concentration. After addition of 1.00 mL of 2.00 x 10-2 M NaCl, the potential changed to -0.1994 V. Calculate the Na+ concentration of the original solution.

Is there a specific equation I can use so I can correctly solve for the concentration of the unknwon solution?

$\endgroup$
2
$\begingroup$

The little problem with the standard addition in potentiometry is, that the response to the addition is not linear, but logarithmic.

If we consider the simplified Nernst equation in context of parameters $A, B$,

and if we consider $c, c_\mathrm{0}$ as unknown concentration and concentration increment, respectively,

we can follow this derivation:

$$\begin{align} E_\mathrm{1} &= A + B \cdot \log c\\ E_\mathrm{2} &= A + B \cdot \log ( c + c_\mathrm{0} )\\ E_\mathrm{2}- E_\mathrm{1} &= B . \log( 1 + \frac {c_\mathrm{0}}{c} )\\ 10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} &= 1 + \frac {c_\mathrm{0}}{c}\\ c &= \frac {c_\mathrm{0}}{10^{\frac{E_\mathrm{2}- E_\mathrm{1}}{B}} -1} \end{align}$$

where $B = \frac{RT}{nF} \cdot \ln(10)$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.