formic acid [HCOOH]aq <-------------> [HCOO-]aq + [H+]aq initally 0.11 M 0 0 at
ID: 699706 • Letter: F
Question
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
IF U SATISFY WITH THE SOLUTION PLEASE LIKE IT ....................THNKXXXX
Explanation / Answer
show all work and equations
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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.