Acurrent 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

10.063g / 10.00 ml = 1.0063 g/ml which rounds off to 4 sig figs as: 1.006 g/ml