Question

When the difference in pH across the membrane of a glass electrode at 25 degree C is 4.40 pH units, how much voltage is generated by the pH gradient? (F = 96485 C/mol; R = 8.3145 J/mol K) mV (three significant figures) (b) What would the voltage be for the same pH difference at 37 degree C? (F = 96485 C/mol; R = 8.3145 J/mol K) mv (three significant figures) When calibrating a glass electrode, 0.025 m potassium dihydrogen phosphate/0.025 m disodium hydrogen phosphate (see the NIST Buffers table) gave a reading of - 19.3 mV at 30 degree C and 0.05 m potassium hydrogen phthalate buffer gave reading of + 144.2 mV. What is the pH of an unknown giving a reading of + 50.1 mV?

Answer 1

$\Delta$ E = RT/nF * $\Delta$ pH

       = 8.314 J/K/mol *298 *2.303/ 1* 96500 C/mol $\Delta$ pH

       = 0.236 V = 236 mV

The voltage will change 236 mV

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$\Delta$ E = RT/nF * $\Delta$ pH

       = 8.314 J/K/mol *303*2.303/ 1* 96500 C/mol $\Delta$ pH

      = 0.240 V = 240mV

at 37oC voltage difference will be 240mV