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T asked in Science & MathematicsChemistry · 4 years ago

If the Ka of a monoprotic weak acid is 6.1 × 10-6, what is the pH of a 0.28 M solution of this acid?

ph=______

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  • hcbiochem
    Lv 7
    4 years ago
    Favorite Answer

    For a weak acid, HA, the equilibrium is:

    HA <--> H+ + A-

    Ka = [H+][A-]/[HA] = 6.1X10^-6

    Let x = [H+] = [A-], and [HA] = 0.28 - x. Because Ka is small, x will probably be much smaller than 0.28 and so can be ignored. Then,

    Ka = 6.1X10^-6 = x^2 / 0.28

    x = [H+] = 1.3X10^-3 (x is indeed small compared to 0.28, so ignoring it was fine)

    pH = - log (1.3X10^-3) = 2.88

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