Answer to Question #170445 in General Chemistry for heidi

Question

A current of 11 A is applied to 1.25 L of a solution of 0.549 molL−1 aqueous HBr converting some of the H+(aq) to H2(g), which bubbles out of solution.which bubbles out of solution.What is the pH of the solution after 77 minutes? (Assume the volume of solution to be constant.)

Answer 1

n(H⁺) = It/Fz

n(H⁺) = 11A * 77*60c/96485Kl/c*z = 0.527 mol

n1 = 1.25l * 0.549 = 0.686 mol

n2 = 0.686 mol - 0.527 mol = 0.159 mol

c2 = 0.159mol / 1.25l = 0,127 mol/L

pH = -lg(H+) = 0.896

pH = 1