althycholyes Ftnaoz MF + OH arep 62.5 MI O:ob tx こフa7 Determine the pH at the following points in the titration of 10.0 mL of 0.200 M formic acid (HCHO,) with 0.100 M NaOH. O,WOM=X (7a) At the start of the titration 2.04 HСИo, Z nt +0 ph=D[H+JCCHO, cl17yIu-40 -l,7y1u-4 こてと10 91200 -X xa=6,200-x)(l.7x0-4) X= 3.u Yl0-5-17vioY* xlo-Ux-3uv0-5 +1.9x10 -3.4~(0- 5 -4 Tyiu-)41-3,4 Y0-5) -117v10-4 2. Naury (tomi) L0200) = \copo) x\cop0) At the midpoint of the titration At midpomt sb sam HCn 102 tHoO CHO2- → Ho 0 こ Wand O:064 とX 5,88 y1011 67 After 12.0 mL of NaOH has been added
althycholyes Ftnaoz MF + OH arep 62.5 MI O:ob tx こフa7 Determine the pH at the following points in the titration of 10.0 mL of 0.200 M formic acid (HCHO,) with 0.100 M NaOH. O,WOM=X (7a) At the start of the titration 2.04 HСИo, Z nt +0 ph=D[H+JCCHO, cl17yIu-40 -l,7y1u-4 こてと10 91200 -X xa=6,200-x)(l.7x0-4) X= 3.u Yl0-5-17vioY* xlo-Ux-3uv0-5 +1.9x10 -3.4~(0- 5 -4 Tyiu-)41-3,4 Y0-5) -117v10-4 2. Naury (tomi) L0200) = \copo) x\cop0) At the midpoint of the titration At midpomt sb sam HCn 102 tHoO CHO2- → Ho 0 こ Wand O:064 とX 5,88 y1011 67 After 12.0 mL of NaOH has been added
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Question 7b
Ph at The midpoint of the titration. My answer is not matching the study guide answer please show all the equations involved
![althycholyes Ftnaoz MF + OH
arep
62.5 MI
O:ob
tx
こフa7
Determine the pH at the following points in the titration of 10.0 mL of 0.200 M formic acid (HCHO,)
with 0.100 M NaOH. O,WOM=X
(7a)
At the start of the titration
2.04
HСИo, Z nt +0
ph=D[H+JCCHO,
cl17yIu-40
-l,7y1u-4
こてと10
91200 -X
xa=6,200-x)(l.7x0-4)
X= 3.u Yl0-5-17vioY*
xlo-Ux-3uv0-5
+1.9x10
-3.4~(0- 5
-4 Tyiu-)41-3,4 Y0-5)
-117v10-4
2.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f94b6c4-8026-41b5-a2ce-ea2500e29589%2F16872881-6ad0-4528-b332-53427b1f2d49%2Fb38dftp.jpeg&w=3840&q=75)
Transcribed Image Text:althycholyes Ftnaoz MF + OH
arep
62.5 MI
O:ob
tx
こフa7
Determine the pH at the following points in the titration of 10.0 mL of 0.200 M formic acid (HCHO,)
with 0.100 M NaOH. O,WOM=X
(7a)
At the start of the titration
2.04
HСИo, Z nt +0
ph=D[H+JCCHO,
cl17yIu-40
-l,7y1u-4
こてと10
91200 -X
xa=6,200-x)(l.7x0-4)
X= 3.u Yl0-5-17vioY*
xlo-Ux-3uv0-5
+1.9x10
-3.4~(0- 5
-4 Tyiu-)41-3,4 Y0-5)
-117v10-4
2.
![Naury
(tomi) L0200) = \copo)
x\cop0)
At the midpoint of the titration
At midpomt
sb sam
HCn
102 tHoO
CHO2-
→ Ho 0 こ
Wand
O:064
とX
5,88 y1011
67
After 12.0 mL of NaOH has been added](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F3f94b6c4-8026-41b5-a2ce-ea2500e29589%2F16872881-6ad0-4528-b332-53427b1f2d49%2Fh79uzh7.jpeg&w=3840&q=75)
Transcribed Image Text:Naury
(tomi) L0200) = \copo)
x\cop0)
At the midpoint of the titration
At midpomt
sb sam
HCn
102 tHoO
CHO2-
→ Ho 0 こ
Wand
O:064
とX
5,88 y1011
67
After 12.0 mL of NaOH has been added
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