Question

22. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.3567 M NaOH? $K_a = 1.8 \times 10^{-4}$ for formic acid. A) 2.06 C) 8.52 B) 5.48 D) 11.94

Name: 22 what is the approximate ph at the equivalence point of a weak acid strong base titration if 25 ml of aqueous formic acid requires 2980 ml of 03567 m naoh ka 18 times 10 4 for formic acid 88288
Uploaded: 2023-04-16T16:44:33-08:00
Duration: 3 min 10 s
Channel: Madhur L
Description: 22 what is the approximate ph at the equivalence point of a weak acid strong base titration if 25 ml of aqueous formic acid requires 2980 ml of 03567 m naoh ka 18 times 10 4 for formic acid 88288

          22. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.3567 M NaOH? $K_a = 1.8 \times 10^{-4}$ for formic acid.
A) 2.06
C) 8.52
B) 5.48
D) 11.94

22 what is the approximate ph at the equivalence point of a weak acid strong base titration if 25 ml of aqueous formic acid requires 2980 ml of 03567 m naoh ka 18 times 10 4 for formic acid 88288

Added by Jose Antonio R.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert Madhur L

04/16/2023

Step 1

Moles of NaOH = Volume of NaOH (in L) $\times$ Molarity of NaOH Moles of NaOH = $29.80 \times 10^{-3} L \times 0.3567 mol/L = 0.01062 mol$ Show more…

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22. What is the approximate pH at the equivalence point of a weak acid-strong base titration if 25 mL of aqueous formic acid requires 29.80 mL of 0.3567 M NaOH? $K_a = 1.8 \times 10^{-4}$ for formic acid. A) 2.06 C) 8.52 B) 5.48 D) 11.94