A current of 10.4 a is applied to 1.25 l of a solution of 0.552 m hbr converting some of the h + to h2 (g), which bubbles out of solution.

The question is incomplete:

Complete question is read as:

A current of 10.4 a is applied to 1.25 l of a solution of 0.552 m hbr converting some of the h + to h2 (g), which bubbles out of solution. What is the pH of the solution after 69 minutes?

...

Solution:

Initial number of moles of H + ions = molarity of solution X volume of solution

= 0.552 X 1.25

= 0.69

Now, during electrochemical reaction conc. of H + ions will decrease in solution.

Using Faraday's law, we know that,

number of moles of H (+) discharged at the electrode = n = Q / (z•F)

where, Q = total electric charge in Coulomb

F = Faraday constant = 96485 C mol-1

z = electrons transferred per ion = 1 (in present case)

We also know that, Q = I x t = 10.4 X 69 X 60 = 43056 C

∴ n = Q / (z•F)

= (43056) / (1 X 96500) = 0.4462

Now, number of moles of H + ions left in solution = 0.69 - 0.4462 = 0.2438

Now, Molarity of solution = number of moles/volume of solution

= 0.2438 / 1.25

= 0.195 M

Finally, conc. of [H+] = 0.195 M

∴ pH = - log [H+] = - log[0.195] = 0.71