Hi, i'm stuck on part b) of the question below:

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?

*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?

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