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If one were to mix 200.0 mL of an HCl solution that had a pH of 3.00 with 200.0 mL of an NaOH solution that had a pH of 10.00, what would be the pH of the final solution?
Can you show me the steps please so that I can apply to questions that ask for the same type of solving? thank you!

  • Chemistry -

    pH = -log (H^+)
    pH = 3.00, (H^+) = 1E-3

    pH = 10.00
    (H^+) = 1E-10 and OH^- = 1E-4

    moles = M x L
    For HCl = 1E-3 x 0.200 = 2E-4 moles.
    For NaOH = 1E-4 x 0.200 = 2E-5 moles.

    .............HCl + NaOH ==> NaCl + H2O
    initial..... 2E-4..2E-5.......0....0
    change......-2E-5.-2E-5....+2E-5 .+2E-5
    So the final solution contains an excess of 1.8E-4 moles HCl, no NaOH, and NaCl and H2O formed don't change the pH, in a final volume of 400 mL (0.400 L).
    (H^+) = moles/L = 1.8E-4/0.400 = 4.5E-4
    pH = -log(H^+) = -log(4.5E-4) = -(3.35) = 3.35.

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