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

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Chapter1: Chemical Foundations
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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
Transcribed Image Text: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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