Calculate the pH of 0.45 M benzoic acid, C6H5COOH. The Ka of benzoic acid is 6.5 x 10-5.
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Answer:
C6H5COOH ⇌ C6H5COO- + H+
Ka=
products
reactants
Ka=
[H+][C6H5COO-]
[C6H5COOH]
=6.3×10-5
[x][x]
[0.40]
=6.3×10-5
√
x2
=
√
(6.3×10-5)(0.40)
x = 5.020x10-3 M = [H+]
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