Question

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The carabid beetle Galerita lecontei has a pair of abdominal defensive glands that secrete an aqueous solution comprised predominantly of formic acid. It is estimated that the gland contains 0.11M formic acid.What is the approximate pH of the glandular

solution assuming it is solely 0.11M formic acid in water. The Ka for formic acid is 1.78 x 10^-4 M.

ANSWER: pH = 2.35

Answer 1

formic acid

[HCOOH]aq <-------------> [HCOO-]aq + [H+]aq

initally 0.11 M 0 0

at eqbm. 0.11-0.11x 0.11x 0.11x

NOW Ka = [HCOO-]*[H+]/[HCCOH]

PUTTING THE VALUES

1.78*10^-4 = [0.11x][0.11x]/[0.11-0.11x]

1.78 * 10^-4 = [0.11x][0.11x]/0.11[1-x]

1.78 * 10^-4 = 0.11x^2/1 as Ka value is very small . so x will be very less 1-x = 1

x^2 = 1.6182 * 10^-3

x= 0.0402266

now the amount o H+ in aqueous solution = 0.11x

=0.11 * 0.0402266 M

= 4.424926 * 10^ -3 M

now pH = -log(H+)

pH = -log(4.424926 * 10^-3)

pH = 2.3540938

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