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ElectroChemistry PH Question

  1. Jul 27, 2011 #1
    1. The problem statement, all variables and given/known data
    A current of 10.7 Amps is applied to a 1.30 L solution of 0.550M HBr converting some of the H+ to H2(g), which bubbles out of solution. What is the PH of the solution after 77 minutes.

    2. Relevant equations

    3. The attempt at a solution
    (10.7 C/s)(1 mol e/96485 C)(2 mol H+/2 mol e)(60s/min)(77 min) = .512 mol H+

    .512 mol H+/1.30L = .394M H+
    -log(.394M) = .405 = PH

    This was incorrect
  2. jcsd
  3. Jul 27, 2011 #2
    You started with a certain amount of H+ in the solution. The electrolysis removed some. Now you need to calculate what's left.
  4. Jul 27, 2011 #3
    I see will try, thank you
  5. Jul 27, 2011 #4
    It worked, thank you
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