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Standard XII

Chemistry

Amino Acids

A current of ...

Question

A current of 10.0A is passed through 1.0 L of 1.0 M HCl solution for 965 seconds pH of the solution at the end of the experiment is

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Solution

1L of 1.0 M HCL solution contain
=1molH+ + e- =>1/2H2
So 1Faraday produce 1/2 mol H2 =0.5mol H2
96510/96500 Faraday produce = 965100.5/96500
=0.05mol H2
From above equation 0.05 mol H2 contain 0.1 mol H+ ion
So H+ ion left in solution=1-0.1=0.9mol
=>therefore pH = -log[H+]=-log(0.9)= --0.0457
=0.0457
pH of the solution at the end of the experiment =0.0457

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