# Glass Electrode calibration and pH analysis Help.

Discussion in 'General Science' started by lam58, Dec 28, 2015.

1. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
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: Dec 29, 2015
2. ### wayneh Expert

Sep 9, 2010
12,120
3,046
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.

3. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
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.

4. ### BR-549 Well-Known Member

Sep 22, 2013
1,987
388
Sounds like a homework problem.

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

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

For that probe.

5. ### wayneh Expert

Sep 9, 2010
12,120
3,046
I don't see how that is right for estimating E0. I admit it's been a long time since I did this sort of stuff.

6. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
Ok, any idea what you think it might be, or rather how to go about find it.

7. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
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.

8. ### wayneh Expert

Sep 9, 2010
12,120
3,046
Try plotting E vs pH

9. ### lam58 Thread Starter Member

Jan 3, 2014
69
0

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?

10. ### wayneh Expert

Sep 9, 2010
12,120
3,046
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?

11. ### lam58 Thread Starter Member

Jan 3, 2014
69
0
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.

12. ### BR-549 Well-Known Member

Sep 22, 2013
1,987
388
What was part A of the question?

13. ### lam58 Thread Starter Member

Jan 3, 2014
69
0

It was asking what was the sensitivity of the device which I figured out to be 0.0562 V.