Chemistry

Consider 51.5 mL of a solution of weak acid HA (Ka = 1.00 10-6), which has a pH of 4.210. What volume of water must be added to make the pH = 5.000

I've tried everything but I am not getting the right answer!

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asked by M
  1. I would do this.
    ........HA --> H^+ + A^-
    I.......x......0.....0
    C......-y......y.....y
    E .....x-y.....y.....y

    You know y = 6.17E-5
    Substitute and solve for x =initial (HA) and let's call that M1. (This will be quadratic you must solve.)

    Do the same for HA in which y = 1E-5 and solve for HA and let's call that M2.

    Then 51.5 x M1 = mL x M2. (another quadratic)
    Solve for mL which will be the final volume, then mL-51.5 will be how much H2O must be added.

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