a.) Formic acid (HCOO H , Ka = 1.77 x 10 -4 ) is the active ingredient in ant ve
ID: 502714 • Letter: A
Question
a.) Formic acid (HCOOH, Ka = 1.77 x 10-4) is the active ingredient in ant venom. Write the dissociation reaction for an aqueous solution of formic acid. ___________________________________________________________________________________________________________________.
b.) Is formic acid a weak acid or a strong acid? what does this mean for the reaction in 4a? ___________________________________________.
c.) Calculate the pH and pOH of a 0.25 M solution of formic acid. Is this solution acidic, basic or neutral?
pH=_________ pOH=__________ acidic,basic or neutral=____________
d.) A solution of formic acid has a pH of 3.00. What is the (initial) concentration of formic acid in this solution?
[HCOOH]0=______________
Explanation / Answer
a) HCOOH + H2O <--------------> HCOO- + H3O+
b) formic acid is weak acid
as Ka is small it is considered as weak acid
so we are using equilibrium arrow in dissociation
c) as it is weak acid
pH = 1/2 [pKa - logC]
pKa = - log Ka = - log [1.77 x 10-4]
pKa = 3.75
pH = 1/2 [3.75 - log 0.25]
pH = 2.18
pOH = 14 - 2.18
pOH = 11.82
solution is acidic in nature
d) pH = 1/2 [pKa - logC]
3.0 = 1/2 [3.75 - logC]
6.0 = 3.17 - logC
logC = -2.83
C = 0.0015 M
[HCOOH] = 0.0015 M
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.