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. a. Write the K₂ expression for formic acid. 1.77x10-4 x K₂=1.0×10-14 Kb=5₂6x10-11 ках ко Kax K₂=1.0×·10 [AC] [H] THCOOHIL 10.006600.000 [0.243.82 1/2 D ka= x³ b. Calculate the pH of a 0.25 M solution of formic acid. 1.77x10-4 (0.25M) [H+]] [H+]=7.08×10²4 PH=-10g (7.08x104) PH=3.15 c. What is the % dissociation of the formic acid? HCOOH HrOt+ CH =1.77x10-4 1.82X10-4-97.25 100-9729-2.75% x² 00065 .25 177x6-4 X100

Introduction to General, Organic and Biochemistry
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Chapter6: Solutions And Colloids
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Problem 6.25P: 6-25 A small amount of solid is added to a separatory funnel containing layers of diethyl ether and...
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CH3NH₂ (aq) + H₂O (1) OH +
CH3NH3
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.
a. Write the K₂ expression for formic acid.
1.77x10-4 x K₂=1.0×10-14
kb=5₂6×10-11
1/2
x²
b. Calculate the pH of a 0.25 M solution of formic acid.
1.77x10-4 = (0,25M) [H + ]
[H+]=7.08×10-4
ka =
R HCOOH
x x
Kaxk₁= 10×·10-14
ках.
[CO] [OH-]
애]
THCOOHIL
10.00660][0.00665
[0.243] 1.82×10²4
PH=-10g (7.08x104)
PH=3.15
-1.77x10-4
11.82x10 -4 = 47.25
100-9725-2.75%
00665
.25
0₂25-1-77x10-4
x²=4.425x10-5
x=0.00665 [H-0005
X [HCCCHI-0.243 [HCO]=-0.00665
2
c. What is the % dissociation of the formic acid?
X
0.25 (0) (0) (09) 0.25
HCO+ + OH
(aq)
O 0
(aq)
-X
X
È 0.25-x
X100
Transcribed Image Text:CH3NH₂ (aq) + H₂O (1) OH + CH3NH3 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. a. Write the K₂ expression for formic acid. 1.77x10-4 x K₂=1.0×10-14 kb=5₂6×10-11 1/2 x² b. Calculate the pH of a 0.25 M solution of formic acid. 1.77x10-4 = (0,25M) [H + ] [H+]=7.08×10-4 ka = R HCOOH x x Kaxk₁= 10×·10-14 ках. [CO] [OH-] 애] THCOOHIL 10.00660][0.00665 [0.243] 1.82×10²4 PH=-10g (7.08x104) PH=3.15 -1.77x10-4 11.82x10 -4 = 47.25 100-9725-2.75% 00665 .25 0₂25-1-77x10-4 x²=4.425x10-5 x=0.00665 [H-0005 X [HCCCHI-0.243 [HCO]=-0.00665 2 c. What is the % dissociation of the formic acid? X 0.25 (0) (0) (09) 0.25 HCO+ + OH (aq) O 0 (aq) -X X È 0.25-x X100
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