How would you solve b? 3. Formic acid, HCOOH (K₂ = 1.77 x 104) is naturally found in the venom of bee stings, and is usedsynthetically as a preservative in livestock feed.Kax k-10x1014b[COF] [OH-][HCOOH]|b. Calculate the pH of a 0.25 M solution of formic acid. 10.00660][0.00665tz[0.243] 1.82×10²41.77x10-4 = (0.25M) [HT][H+] = 7.08 × 10 -4pH = -10g (7.08×104)PH=3.15a. Write the K, expression for formic acid.1.77x10-4 x K₂=1.0×10-14Kb=5₂6×10-11c. What is the % dissociation of the formic acid?+= 0.25M (ag) !-X0.25-xXXka =каDOLLS25X0.25 = 1-77x10-4x² = 4.425x10-5x=0.00665 1=00005X [HCCCH]=0.243 [HCO]=000660X0=1.77x10-41.82x10-4-97.25100-97.29-2.75%X100

Given :- Molarity of Formic acid, HCOOH = 0.25 M Ka for Formic acid, HCOOH = 1.8 × 10-4 pH of…

Answered: b. Calculate the pH of a 0.25 M…

b. Calculate the pH of a 0.25 M solution of formic acid. 1.77x10-4-(0.25M) [H+] [H+]=7.08×10-4 pH = -10g (7.08×10-4) PH=3.15

Chemistry: The Molecular Science

5th Edition

ISBN:9781285199047

Author:John W. Moore, Conrad L. Stanitski

Publisher:John W. Moore, Conrad L. Stanitski

Chapter15: Additional Aqueous Equilibria

Section: Chapter Questions

Problem 124QRT

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Question

How would you solve b?

3. Formic acid, HCOOH (K₂ = 1.77 x 104) is naturally found in the venom of bee stings, and is used
synthetically as a preservative in livestock feed.
Kax k-10x1014
b
[COF] [OH-]
[HCOOH]|
b. Calculate the pH of a 0.25 M solution of formic acid. 10.00660][0.00665
tz
[0.243] 1.82×10²4
1.77x10-4 = (0.25M) [HT]
[H+] = 7.08 × 10 -4
pH = -10g (7.08×104)
PH=3.15
a. Write the K, expression for formic acid.
1.77x10-4 x K₂=1.0×10-14
Kb=5₂6×10-11
c. What is the % dissociation of the formic acid?
+
= 0.25M (ag) !
-X
0.25-x
X
X
ka =
ка
DOLLS
25
X
0.25 = 1-77x10-4
x² = 4.425x10-5
x=0.00665 1=00005
X [HCCCH]=0.243 [HCO]=000660
X
0
=1.77x10-4
1.82x10-4-97.25
100-97.29-2.75%
X100

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