Glass Electrode calibration and pH analysis Help.

Thread Starter

lam58

Joined Jan 3, 2014
69
Hi, i'm stuck on part b) of the question below:

Q: In order to perform pH determinations with a glass electrode, the cell potential was measured for threestandard solutions with the following pH values at 25 Celsius: 2.04, 7.05, and 9.20. The cell voltage readout(in mV) for each of the above solutions was 238.0, -37.5 and, -164.5, respectively. Calculate: (a) thesensitivity of the pH sensor; (b) the pH of an unknown sample yielding a cell voltage of 20.5 mV; (c) thepH deviation from the actual value if the sample temperature is 35 Celsius.

My ans:
So to get the sensitivity (s) I just use the nernst equation i.e.

\( s = \frac{2.303 RT}{nF}\)


Where: R is the gas constant \( 8.3144 J.K^{-1}.mol^{-1}\)
T is temperature in Kelvins
n is the charge of a Hydrogen ion = 1
F is Faraday's constant \( 96485 C.mol^{-1}\)

\(\Rightarrow \frac{2.303 \times 8.3144 \times 298}{96485} = 0.0591 \)V/pH unit


But now I'm stuck on part b). My method is either to rearrange the nernst equation like so:

\( E = E^0 + \frac{2.303 RT}{nF}. pH\)

\(\Rightarrow pH = \frac{E - E^0}{2.303 RT/ nF}\)

where \(E^0 = 238mV - 37.5mV - 164.5mV = 36mV\)

\( = \frac{0.0205v + 0.036v}{0.0591 v/pH unit}\)


But this gives me a pH value of 0.96 which doesn't seem right.

The other method I am using is to work out the value given the pH scale and sensitivity. i.e.
If we know that the cell voltage of the unknown pH is 0.0205V, then if we find the difference between this and the the voltage for pH 7.05 (0.0375V) gives 0.058V. Then divide by the sensitivity of 0.0591V/pH unit to give a pH difference of 0.981. Then then take this value away from the reference pH value of 7.05 to give a pH of 6.068.
Any thoughts on my methods, are they correct or am I way off?
 
Last edited:

wayneh

Joined Sep 9, 2010
16,399
The two higher pH values give negative voltages, correct? (I see an accented "A").

By inspection you can see the pH needs to be between the first two references and much closer to the 7 than the 2. So a 6 makes more sense than a 1.

Try using units in your calculations. It would make them easier to follow and perhaps your error would become obvious.
 

Thread Starter

lam58

Joined Jan 3, 2014
69
The two higher pH values give negative voltages, correct? (I see an accented "A").

By inspection you can see the pH needs to be between the first two references and much closer to the 7 than the 2. So a 6 makes more sense than a 1.

Try using units in your calculations. It would make them easier to follow and perhaps your error would become obvious.
Hi, sorry about that, I posted this late last night and was pretty tired. I've edited the above to include units. Still, I'm not sure what I'm doing wrong. I reckon the answer to my second method to find the pH for a cell voltage of 0.0205v sounds right, but I may have just found it by a chance. I'm really not sure. The problem is all the books I've looked at don't really explain it very well either.
 

BR-549

Joined Sep 22, 2013
4,938
Sounds like a homework problem.

pH probes come with calibration sheets. At least they use to.

Google the probe number and download sheets.

These sheets have plots of the relationships and show the relevant equations.

For that probe.
 

Thread Starter

lam58

Joined Jan 3, 2014
69
Sounds like a homework problem.

pH probes come with calibration sheets. At least they use to.

Google the probe number and download sheets.

These sheets have plots of the relationships and show the relevant equations.

For that probe.
It's not homework it's a tutorial question to help me understand.

No it's not a real glass electrode it's just a question about the principles of glass electrodes and how to analyse them, which of course I am failing at.
 

Thread Starter

lam58

Joined Jan 3, 2014
69
Try plotting E vs pH

Hi again thank for the reply. So I tried plotting everything and for question 1a) I got the sensitivity to be -0.0562 V. i.e.

if,
\( y = mx + b\)

\( \Rightarrow m = \frac{dy}{dx} = \frac{(0.238 - - 0.0375)+(0.238 - -0.1645)+(0.1645-0.0375)}{(9.2-2.04)+(9.2-7.05)+(7.05-2.04)}\)

\( = \frac{0.805}{14.32} = -0.0562 V/pH unit\)

For b) to work out the pH of the a cell voltage of 0.0205 V, I first worked out what the y intercept of the graph:

\( y=mx + b \Rightarrow b = y - mx\)

\( \Rightarrow b = 0.238 - (-0.0562 \times 2.04) = 0.353\)

Then I rearranged the graph equation to find what x (pH) would be if y = 0.0205 V:

\( x = \frac{y - b}{m} = \frac{0.0205 - 0.353}{-0.0562} = 5.91\)

For part c) I used the Nernst equation to find the difference in cell voltage output when the temperature has at 35 celsius then worked out the pH deviation:

\( E = E^0 - \frac{2.303 RT}{nF} \times pH = 0.353 - (\frac{2.303 \times 8.3144 \times (273+35)}{1 \times 96485} \times 5.91)\)

\( = 0.353 - 0.361 = -0.008\)

This implies that the cell voltage at 35 celsius is:

\( 0.0205 - 0.008 = 0.0125 V\)

and pH would be:

\( \frac{0.0125 - 0.053}{-0.0562} = 6.041\)

Giving a deviation of \(\frac{6.041 - 5.91}{35^o C - 25^o C} = 0.0131 pH units/ ^o C\)

Is this right?
 

wayneh

Joined Sep 9, 2010
16,399
It sounds like you've got it now.

So in the plot, did you get a straight line? You always do with 2 points, but rarely do with 3! (If it's real data.) How did you deal with the non-linearity?
 

Thread Starter

lam58

Joined Jan 3, 2014
69
It sounds like you've got it now.

So in the plot, did you get a straight line? You always do with 2 points, but rarely do with 3! (If it's real data.) How did you deal with the non-linearity?
I got a rough straight line with the 3 calibration values hence when working out dy/dx I used all of the values. However I'm not quite sure how to adjust for non-linearity, should I just use two values instead? There seems to be some error when I use the m and b values for the y = mx + b equation that I found. i.e. when I put the values in to calculate the cell voltage for the pH of 7.05 I get -0.043V instead of -0.0375V.
 
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