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Homework Help: Glass electrode voltage calibration and pH analysis.

  1. Dec 28, 2015 #1
    • Please post this type of questions in the HW section using the template.
    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 C: 2.04, 7.05, and 9.20. The cell voltage readout(in mV) for each of the above solutions was 238.0, À37.5 and, À164.5, 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 C.

    My ans:

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

    [tex] s = \frac{2.303 RT}{nF}

    \Rightarrow \frac{2.303 \times 8.3144 \times 298}{96485} = 0.0591[/tex]

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

    [tex] \frac{0.0205 + 0.036}{0.0591}[/tex]

    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?
  2. jcsd
  3. Dec 29, 2015 #2


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    Staff: Mentor

    Have you heard about linear regression? 0.0591 V is a theoretical electrode response, real electrodes almost always deviate.
  4. Dec 29, 2015 #3
    Hi, Thanks for the reply. OK, so does that mean I've done something wrong here? I'm sure I was supposed to work out the theoretical electrode response in part a) of the question? What about part b), any suggestions?
  5. Dec 29, 2015 #4


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    Staff: Mentor

    If you are to determine pH, theoretical response doesn't matter, you do the calibration to find out what the real response is.
  6. Dec 29, 2015 #5
    Sorry, I'm confused by what you mean?
  7. Dec 29, 2015 #6


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    Staff: Mentor

    You should not assume electrode response is ideal. You do the measurements to find out the real, experimental slope of the calibration line, and you use the calibration line to convert the reading (in mV) to pH.
  8. Dec 29, 2015 #7
    OK, that seems to give me the same answer as i got in my second method. :frown:
  9. Dec 29, 2015 #8


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    Staff: Mentor

    Then you are doing something wrong. Have you tried to make a simple plot? My bet is that "À" stands for minus.
  10. Dec 31, 2015 #9

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

    [tex] y = mx + b[/tex]

    [tex] \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)}[/tex]

    [tex] = \frac{0.805}{14.32} = -0.0562 V/pH unit[/tex]

    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 is:

    [tex] y=mx + b \Rightarrow b = y - mx[/tex]

    [tex] \Rightarrow b = 0.238 - (-0.0562 \times 2.04) = 0.353[/tex]

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

    [tex] x = \frac{y - b}{m} = \frac{0.0205 - 0.353}{-0.0562} = 5.91[/tex]

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

    [tex] 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)[/tex]

    [tex] = 0.353 - 0.361 = -0.008[/tex]

    This implies that the cell voltage at 35 celsius is:

    [tex] 0.0205 - 0.008 = 0.0125 V[/tex]

    and pH would be:

    [tex] \frac{0.0125 - 0.053}{-0.0562} = 6.041[/tex]

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

    Is this right?
  11. Jan 1, 2016 #10


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    It seems to be right enough that it is with getting it quite right. Have a care for the final reader (and also us) making sure THAT what the strings of numerical calculation good correspond to are quite obvious to him. It will certainly be helpful to you, to us, and to the final reader, to see your graph, with its three measurement points Plus the one in your pH calculation which last should by the way you have done it fall exactly on the line.

    What you have done here is not what Borek suggested and not really sound. In fact you have effectively taken the difference between the first and last potential and divided it by the difference between the first and last pH: (y1 - y3)/(x1 - x3) - check out the cancellations in your formulae. (So I'll bet that, unless the linearity is very good, you have a line that goes exactly through the outside points and the middle one is slightly off the line. This would have been obvious to us and that there was something slightly wrong if we'd seen the diagram). It would be better to draw what seems to you the best linea fit with ruler, hand and eye through the points. Better still learn about least-squares fitting, or use an online or hand calculator routine to do it for you.

    Numerically in this case it's probably not going to make it any big difference but you need to get the principle right. It looks like the slope it Is quite close to the theoretical one of 0.059, which fact you should mention in your writeup.

    I'll check it out if there's anything to say on the rest of your document later.
    Last edited: Jan 1, 2016
  12. Jan 2, 2016 #11
    Hi, thank you very much for your help. So I tried the above with linear regression and on my calculator and I got pretty much the same answer for m and b like you predicted, however, it is still useful to know. Thanks again.
  13. Jan 10, 2016 #12


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    I am sorry I have taken so long to get back on this problem. A couple of times I did look at it I found your numbers a bit confusing.

    We have seen the slope of the graph is near to ideal, a fact worth noting. The intercept depends on the reference calomel or whatever reference electrode - I don't know if anyone has anything to say about that.

    For question c, better to state what it is as otherwise it might be confusing when you come back to it later. It is asking: if you had calibrated that electro that 25°C and then measured the potential without corrections, what would you think the pH was? Of course warming a solution of for example dilute strong acid changes the potential but cannot change [H+]. I cannot follow your calculations quite; it seems to me from the formulae that the new pH must simply be the old multiplied by the ratio of the two absolute temperatures with R and F not needed which is 1.03 giving an apparent pH of 6.11 approximately, in the same ball park but different from yours. Other opinions to sort this out would be welcome.

    These calculations you should state are subject to the proviso that heat of ionisation of substances in the solution are low enough to not interfere. That is, the pK and hence the pH of buffers will change with temperature, so if you don't want to complicate the interpretation of temperature effects on phenomena (e.g. ligand binding) it is is well to choose a buffer substance with low heat of ionisation. Otherwise what you think is a temperature effect might just be a pH effect! Among standard buffers MOPS is one such, phosphate is another. I think it would be enlightening for you to calculate the real pH change for a 10° rise in temperature of the lowest and highest ΔH0 for say 1M, 0.1 M and 0.01M buffers when pH = pKa .

    Here tables of RT/F, http://www.horiba.com/application/m...-water-quality/the-basis-of-ph/meaning-of-ph/
    and here of ΔH0's, http://chemwiki.ucdavis.edu/Physica...ale/Temperature_Dependence_of_pH_in_Solutions
    both inside some useful accounts
    Last edited: Jan 10, 2016
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