Calibration of a glass electrode gave a reading of 140.1 mV with 0.05 m potassiu

Question

Calibration of a glass electrode gave a reading of 140.1 mV with 0.05 m potassium dihydrogen citrate buffer standard (pH = 3.766) and a reading of -75.9 mV with 0.008695 potassium dihydrogen phosphate, 0.0343 m disodium hydrogen phosphate buffer standard (pH = 7.400), both measured at 30°C.

(a) What is the observed slope (mV/pH unit) of the calibration curve?

(b) What is the theoretical slope at 30°C?

(c) What is , the electromotive efficiency to three significant figures?? Open the hint panel below for the definition of electromotive efficiency.

(d) What is the pH of an unknown that gives a reading of 16.8 mV with this electrode at 30°C?

Please help me out...I am not sure how to start or approach this problem. I am studying for a test and I would like to see how it is solved. Help is greatly appreciated :)

Thank you!

Explanation / Answer

a) E = const + B(0.059) log H+   pH = -log H+ so:

E = const - B(0.059) pH

Now, given pH1 = 3.766 ---> E1 = 140.1 mV

Given pH2 = 7.4 ---------> E2 = -75.9 mV

Observed slope = E2 - E1 / pH2 - pH1

Observed slope = -75.9 - 140.1 / 7.4 - 3.766

Observed slope = -59.44

b) Theorical slope:

E(T) = E°(T) - 0.1984(T) pH

E(T) = -0.1984 x T

E(T) = -0.1984 * (30+273)

E(T) = -60.1152

c) Electromocy efficient Beta:

B = observed slope / theorical slope

B = 59.44 / 60.11

B = 0.9888

d) slope = E2 - E1 / pH2 - pH1

-59.44 = -75.9 - 16.8 / 7.4 - pH1

-59.44(7.4 - pH1) = -92.7

-439.856 + 59.44pH1 = -92.7

pH1 = 5.84

Hope this helps

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