# Analysis of Cations and Anions by Ion-Selective Electrodes (ISEs)


Q1:  Will a more acidic sample displace more, the same or less Na+ from the hydrated gel layer?

Because H+ binds more strongly than Na+ to the –SiO- function in glass, the H+ will displace more Na+.   As the sample becomes more acidic, there are more H+ present.  This greater H+ concentration will displace more Na+ from the hydrated gel.

Q2:  What do you think is meant by mobility of ions?

Students will usually consider mobility to mean how well something moves around.  The mobility of ions can be defined as the rate at which ions move in an applied electric field.

Q3:  In a hydrated membrane, which ion do you think has a higher mobility, H+ or Na+

The smaller an object is the more mobile it will tend to be.  For ions, the mobility is inversely proportional to their effective ionic diameter.  In a hydrate membrane the H3O+, being larger than Na+, will have a smaller mobility.

Q4:  Do you think other cations (e.g., Li+. K+) may have some ability to migrate into the hydrated gel layer of a pH electrode?  If so, is this a problem?

Small cations such as Li+, Na+, and K+ can migrate into the gel layer replacing hydrogen ions and giving rise to a higher membrane potential.  Hydrogen ions are replaced with sodium ions (decreasing the hydrogen ion activity), thereby artificially suppressing the true pH value.  Since lithium ions are rarely present in solution and interference from the potassium ion is minimal due its larger size, sodium ions present the largest interference.  This effect is therefore known as the sodium error in pH measurements.

Q5:  Consider a solution that has some Na+ and very high concentrations of K+Cl-.  What effect do you think this might have on the activity of Na+ in the solution?

The high concentration of K+Cl- will affect the ionic strength of the solution, and, in turns, affect the activity coefficient of Na+.    This effect can be calculated more formally using the extended Debye-Huckel equation (Equation 6.63 in the Harvey Analytical 2.0 text).

Q6: If the indicator electrode potential under standard conditions is -0.100 V, what is the indicator electrode potential at 298 K if the activity of the sodium ion is 0.10 M?

To answer this question, we can use the Nernst equation as in (4):

$\mathrm{E_{ind}= E_{ind}^o- \dfrac{RT}{nF} \ln\left(\dfrac{1}{a_{Na}} \right)} \tag{4}$

where E° is the indicator electrode potential under standard conditions (298 K, 1.00 M Na+), R is the molar gas constant (8.314 J K-1 mol-1) , T is the absolute temperature (K) , n is the number of moles of electrons in the half-reaction, and F is Faraday’s constant (96485 C mol-1). In this particular case

$\mathrm{E_{ind}= -0.100\:V- \dfrac{ \dfrac{8.314\:J}{K\:mol}\times 298\:K}{1\:mol\: e^- \times \dfrac{96485\:C}{mol}} \ln ⁡\left(\dfrac{1}{0.1\:M}\right)}\nonumber$

Therefore Eind = -0.1591 V

Q7: How does the indicator electrode potential change in the previous question if the temperature is increased by 10 degrees?

$\mathrm{E_{ind}= -0.100\:V- \dfrac{\dfrac{8.314\:J}{K\:mol}\times 308\:K}{1\:mol\: e^- \times\dfrac{96485\:C}{mol}} \ln⁡\left(\dfrac{1}{0.1\:M}\right)}\nonumber$

Therefore Eind = -0.1611 V

Thus, the measured potential is dependent on the temperature.  Because of this temperature effect, most meters have an automatic temperature correction.

Q8:  How would you go about calibrating a sodium ion selective electrode?

In order to calibrate the electrode, its response to changes in concentration is measured and recorded.  There are several practical ways to accomplish this.  Most ISEs are standardized by the standard addition method, where a known volume of distilled water and ionic strength adjuster is spiked with increasing concentrations of a standard solution of the analyte. The potential after each addition is recorded.  The calibration is constructed by plotting the electrode potential as a function of the logarithmic transformation of the standard concentration at each addition.

The electrode could be also be calibrated using different solutions of known concentration (similar to how the pH electrode is calibrated using at least two solutions of known pH).

Q9: Can you think of a way to mitigate possible effects of ionic strength to insure that your calibration procedure and sample analysis provide an accurate measurement of the concentration of Na+ in the unknown?

The ionic strength needs to be rather constant, so the ionic strength is adjusted using the addition of an ionic strength adjuster solution.  This solution contains a high concentration of ions that do not interfere or mask the indicator electrode response.

Q10: In the potentiometric determination of sodium ion of a mineral water sample, indicate if either of the following supporting electrolytes can be used for ionic strength adjustment: a 4.0M NH3 – NH4Cl buffer (pH 10) or 4.0M NaCl.

The 4.0 M NaCl would introduce sodium ions into the solution, which affects the concentration of what we are trying to measure.  Therefore the ammonia buffer solution would be a better choice as ionic strength adjuster.

Q11:  What would be the general criteria you would need to use in selecting a suitable supporting electrolyte for an analysis using an ion selective electrode?

One would want a supporting electrolyte that:

1. is different than the analyte.
2. will not interfere with the concentration of the analyte (it does not react or bind with the analyte)
3. it will not mask the presence of the analyte
4. the indicator electrode does not respond to it.
5. it will not foul the electrode or any electrode junction.

Q12:  Based on the relationship in eq 6, how would you construct a calibration that links the changes in electrode potential to changes in the concentration of the sodium ion?

In the Nernst equation (eq 6) the cell potential varies with the log of the activities.  The measured electrode potential is plotted versus the log of the concentration.  The line of best fit is linear.  The equation of the line of best fit is used to determine the concentration of the unknown.

Q13: What is the expected slope of a potentiometric calibration curve for sodium at 35°C? What effect does temperature have on the slope of a potentiometric calibration curve?

$\mathrm{E_{ind}= E^0-\dfrac{\dfrac{8.314\:J}{K\:mol}\times 308\:K}{1\:mol\: e^-\times\dfrac{96485\:C}{mol}} 2.303 \log⁡(Na^+)}\nonumber$

If one plots the potential as a function of logarithm in base ten (log) concentration, then the slope would be 0.0611 since ln x = 2.303 log x

Q14: If a sample has a sodium concentration of 1.0 x 10-3 M, and the sodium ISE has a selectivity coefficient of KNa,H = 30, what sample pH would cause a 1% error in the sodium ISE response?

First calculate the Eind for the solution of 1.0 x 10-3 M

$\mathrm{E_{ind}=-.100\:V- \dfrac{0.05915}{1\:mol} \log\{1\times10^{-3} \}}\nonumber$

$\mathrm{E_{ind}=0.07745\:V}\nonumber$

A 1% error in the response would give a

$\mathrm{E_{ind}=0.07822\:V}\nonumber$

Putting this back into the original equation

$\mathrm{0.07822\:V=-.100\:V- \dfrac{0.05915}{1\:mol} \log\{x\}}\nonumber$

Thus x = 9.70 x 10-4 M which represents the apparent sodium ion concentration due to 1% error.

Putting this into equation 8:

$\mathrm{K_{Analyte,Interferent}=\dfrac{a_{Analyte}}{(a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} }}} \tag{8}$

$\mathrm{9.70 \times 10^{-4}=\{ K_{Analyte,Interferent} (a_{Interferent} )^{n_{Analyte}⁄n_{Interferent} } \}}\nonumber$

$\mathrm{9.70 \times 10^{-4}=\{ 30(a_{Interferent} )^1 \}}\nonumber$

Leads to an H+ concentration of 3.2 x 10-5 M, or pH 4.5

Q15: Evaluate whether it is best to use alkaline or acidic conditions to determine the sodium ion concentration by ISE?

Alkaline conditions are preferred, as in acidic conditions, a higher H+ activity leads to a higher % error.

Q16: The table below contains sodium ISE calibration data. If the cell potential measured in a sample is ‑0.115 V, determine the sodium concentration (mol L-1) in this sample.

[Na+] (M)

Ecell (V vs SCE)

1.0 x 10-4

-0.221

1.0 x 10-3

-0.164

1.0 x 10-2

-0.107

1.0 x 10-1

-0.048

In Excel the graph would look like:

The best fit line is -0.115V = 0.0576x +0.009

The log of Na+ is -2.152

Thus the concentration of Na+  is 7.0 x 10-3 M

Q17: In the previous question, the sample was prepared by pipetting 5.00 mL of the original water sample and 2.00 mL of an ionic strength adjustment buffer into a 100 mL volumetric flask and diluting to the mark with distilled water. Determine the sodium concentration (mol L-1) in the original water sample.

$C_1\times \mathrm{5.00\:mL= 7.0\times 10^{-3}\: M \times 100\:mL}\nonumber$

Thus the original concentration is 0.14M.

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