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Calculate the pH of a 0.39 M CH3COOLi solution. Ka for acetic acid= 1.8 x 10^-5

Calculate the pH of a 0.39 M CH3COOLi solution. Ka for acetic acid= 1.8 x 10^-5

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Answer #1

CH3COOLi is a strong salt, So it dissociates completely

CH3COOLi CH3COO- + Li+

Now,

CH3COO- + H2O CH3COOH + OH-

Ka for acetic acid= 1.8 x 10-5

Kb = Kw/Ka = 1 x 10-14 / 1.8 x 10-5 =5.6 x 10-10

But,

Kb = [CH3COOH] [OH- ] / [CH3COO-]

5.6 x 10-10 = x2 / (0.39 - x )

5.6 x 10-10 = x2 / 0.39 (Neglecting x term from denominator since x << 0.39)

  x2 = 2.184 x 10-10

  x = 1.48 x 10-5

[OH-] = x = 1.48 x 10-5 M

pOH = - log [OH-] = -log(1.48 x 10-5) = 4.83

pH + pOH =14

pH = 14 - pOH = 14 - 4.83 = 9.17

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