Question

# a.) Formic acid (HCOOH, Ka = 1.77 x 10-4) is the active ingredient in ant venom....

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=______________

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