9. How many moles of NaOH would need to be added to 250.0 mL of 0.200 M acetic acid in order to make a buffer solution with a pH of 5.80, assuming that the volume of the solution remains constant?

Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter13: Acids And Bases
Section: Chapter Questions
Problem 29QAP: How many grams of HI should be added to 265 mL of 0.215 M HCI so that the resulting solution has a...
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Question 9 only please. I think it needs a ICE table and a ICF table? Can you use previous info from other questions as well.
9. How many moles of NaOH would need to be added to 250.0 mL of 0.200 M acetic acid in order to
make a buffer solution with a pH of 5.80, assuming that the volume of the solution remains constant?
Transcribed Image Text:9. How many moles of NaOH would need to be added to 250.0 mL of 0.200 M acetic acid in order to make a buffer solution with a pH of 5.80, assuming that the volume of the solution remains constant?
B
с
A
7. Calculate the pH of a buffer made with 50.0 mL of 0.200 M HA and 150.0 mL of 0.100 M NaA, if the Ka
for this generic acid is 2.75x10-4.
0.200 MOHA
1
HA + H₂0 H30² + A
IL
10.050ndl
0.018
-X
0.050-X
с
E
A + HCl -> CI + HA
-3
1.5xi 2.0x13³³
50ML X
HA
0 0.075,0901 150 ML NARX
+ X
+ X
x
0.0750+X
(0.050-X)
X = -(2.75×10² +0.015.0) +√ (2-75410 % 40,075)² + 4(2.75X78) (0.850)
1.822X10M 0.0748M
PH = 3.74
2
- 210X10³ - 2icoxio
4
0.01304,
055
0.0498
ka=-10ka 8. If 2.00 mL of 1.00 M HCl is added t the buffer above, what is the resulting pH? X=1.822224507x10
- 10g (2.75X10") = 3:56 19² [*] -0.0145
19.96x103
-3
4210X16
3.56 +10g
0101196
200,00ML
0.160nolla 1
X
IC
0.01304
0.01196
2.75 X10
= 0.050nd
HA
200.000
(X)(0+0750+x)
020750MB GU
IL
= 0.075
401
Nak
= 3.5975
PH=3.60
-4
0.100 M
31,5x10
9|Page
Transcribed Image Text:B с A 7. Calculate the pH of a buffer made with 50.0 mL of 0.200 M HA and 150.0 mL of 0.100 M NaA, if the Ka for this generic acid is 2.75x10-4. 0.200 MOHA 1 HA + H₂0 H30² + A IL 10.050ndl 0.018 -X 0.050-X с E A + HCl -> CI + HA -3 1.5xi 2.0x13³³ 50ML X HA 0 0.075,0901 150 ML NARX + X + X x 0.0750+X (0.050-X) X = -(2.75×10² +0.015.0) +√ (2-75410 % 40,075)² + 4(2.75X78) (0.850) 1.822X10M 0.0748M PH = 3.74 2 - 210X10³ - 2icoxio 4 0.01304, 055 0.0498 ka=-10ka 8. If 2.00 mL of 1.00 M HCl is added t the buffer above, what is the resulting pH? X=1.822224507x10 - 10g (2.75X10") = 3:56 19² [*] -0.0145 19.96x103 -3 4210X16 3.56 +10g 0101196 200,00ML 0.160nolla 1 X IC 0.01304 0.01196 2.75 X10 = 0.050nd HA 200.000 (X)(0+0750+x) 020750MB GU IL = 0.075 401 Nak = 3.5975 PH=3.60 -4 0.100 M 31,5x10 9|Page
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