Question

A current of strength 96.5 ampere is passed for 10 sec through 1litre aqueous solution of 0.1 miles of copper sulphate calculate ph

Answer 1

Answer:

1

Explanation:

Given A current of strength 96.5 ampere is passed for 10 sec through 1litre aqueous solution of 0.1 miles of copper sulphate calculate ph

Given I = 96.5 A and time = 10 secs

Charge Q = I x t

               = 96.5 x 10  

               =9650 C

We know that Faraday’s constant is 96500 c/mol

Now to decompose 4 moles of OH-, we need 4 moles of e- which will be 4 x 96500 C = 3,86,000 C

So 9650 will decompose in 9650 x 4 / 3,86,000 mole OH-

= 0.1 mole OH-

H+ concentration in solution will be = [0.1] M

So pH = - log [H+] = - log [0.1]

                        = 1