Question
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.)
What is the pH of the solution after 68 minutes? (Assume the volume of solution to be constant.)
Answers
DrBob222
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.
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.