chemistry

What is the Ka of a weak acid (HA) that produces a pH of 3.45 with 0.1M?
2.54E-3
3.45E-3
1.27E-6
3.56E-3
5.48E-4

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asked by treasa
  1. pH = 3.45
    3.45 = -log(H^+)
    (H^+) = approx 3.55E-4.
    ...........HA ==> H^+ + A^-
    I.........0.1.....0.....0
    C..........-x.....x.....x
    E.........0.1-x...x.....x
    where x = 3.55E-4
    Substitute that into Ka expression and solve for Ka.

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