A current of 10.7 A is applied to 1.25 L of a solution of 0.550 molL−1 aqueous HBr converting some of the H+(aq) to H2(g), which bubbles out of solution.
What is the pH of the solution after 68 minutes? (Assume the volume of solution to be constant.)
How many coulombs do we have? That's
C = amperes x seconds = 10.7 x 68 min x (60 sec/min) = 43,656
96,485 coulombs will release 159.8/2 =79.9 g Br2 which means we released
79.9 g Br2 x (43,656/96,485) = 36.2 g Br2 released or
1 g H2 x (43,656/96,485) = 0.452 g H2 or 0.226 moles H2 = 0.452 moles H^+ released . You had 0.550 mols HBr/L x 1.25 L = 0.687 mols HBr initially which is 0.687 mols H^+ initially. You now have 0.687 - 0.452 = 0.235 moles H^+.
M HBr = moles HBr/L solution so assuming no change in volume that is M HBr = 0.235/1.25 = ? and convert that to pH. Check these numbers. It's late and I've been staring into the computer for too long.