What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid require

Question

3 years ago

6

s 29.80 mL of 0.3567 MNaOH? Ka = 1.8×10−4 for formic acid.

2 answers:

iogann1982 [59] · Answer 1 · 2022-08-16T19:30:53+03:00

Answer:

pH= 0.369

Explanation:

The formic acid reacts with NaOH as

HCOOH + NaOH= HCOONa + H2O

Apply CaVa/ CbVb = Na/Nb

Ca×25/(29.8×0.3587) = 1/1

Ca= (29.8×0.3587)/25= 0.428M

pH = - log(H+)

Since only 0ne H+ is in the stoichiometric equation, it means H+ = 0.428M

pH = -log(0.428) =0.369

laiz [17] · Answer 2 · 2022-08-16T17:58:21+03:00

laiz [17]3 years ago

4 0

Explanation:

Below is an attachment containing the solution