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 K, for this generic acid is 2.75x10. 0.200 MOHA 1 HA + H₂0 H30² + A X IL 0.050ndl 1 C E -x 0.050-X 50MLX HA 0 0.075,0pol 150 ML NARX +x +X = 0.050nd 200.00ML 0.160nolak | X 2,75 X10 = 0.075 IE 200.00m - 4 (X) (0+0750+2) Nat 0.0750+X (0.050-x) 0.0498 X = -(2.75×10¹ +0.0150) +√ (2-75410*40675)² + 4(2-7555) (0.850 PH = 3.74 1.822X10M 010748M 2 8. If 2.00 mL of 1.00 M HCl is added to the buffer above, what is the resulting pH? X=1.822224507x10 [HA] = 0+0498 [138] =D=0182 [A] = 0.0748 M
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 K, for this generic acid is 2.75x10. 0.200 MOHA 1 HA + H₂0 H30² + A X IL 0.050ndl 1 C E -x 0.050-X 50MLX HA 0 0.075,0pol 150 ML NARX +x +X = 0.050nd 200.00ML 0.160nolak | X 2,75 X10 = 0.075 IE 200.00m - 4 (X) (0+0750+2) Nat 0.0750+X (0.050-x) 0.0498 X = -(2.75×10¹ +0.0150) +√ (2-75410*40675)² + 4(2-7555) (0.850 PH = 3.74 1.822X10M 010748M 2 8. If 2.00 mL of 1.00 M HCl is added to the buffer above, what is the resulting pH? X=1.822224507x10 [HA] = 0+0498 [138] =D=0182 [A] = 0.0748 M
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 28QRT: Calculate the pH change when 10.0 mL of 0.100-M NaOH is added to 90.0 mL pure water, and compare the...
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![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 2.75×104.
0.200 MOHA
X
IL
P
C
E
50MLX
HA + H₂0 H30⁰ + A HA
0.050nol
0.050-X
0 0.075,0p01 150 ML NARX
+ X
+x
= 0.050nd
HA
200,00ML
0.160nolla 1
2,75 X10
= 0.075
200.00m
IC
-4 (X)(0+0750+1)
=
Nak
x
0.0750+X
(0.050-x)
0.0498
X = -(2.75×10¹ +0.0150) +√(2_15410*40w73)³ +4(2-75x8) (0.850)
PH = 3.74
1.822X10M 0.0748M
2
8. If 2.00 mL of 1.00 M HCI is added to the buffer above, what is the resulting pH? X=1.822224507x10
[HA] = 0+0498 [H38] = 080182 [A] = 0.01481.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F627ffe10-a730-4e4c-b7e9-b306478e5985%2Ff883331a-590a-4b67-b70a-9b467748a856%2Fenbrqm_processed.jpeg&w=3840&q=75)
Transcribed Image Text: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 2.75×104.
0.200 MOHA
X
IL
P
C
E
50MLX
HA + H₂0 H30⁰ + A HA
0.050nol
0.050-X
0 0.075,0p01 150 ML NARX
+ X
+x
= 0.050nd
HA
200,00ML
0.160nolla 1
2,75 X10
= 0.075
200.00m
IC
-4 (X)(0+0750+1)
=
Nak
x
0.0750+X
(0.050-x)
0.0498
X = -(2.75×10¹ +0.0150) +√(2_15410*40w73)³ +4(2-75x8) (0.850)
PH = 3.74
1.822X10M 0.0748M
2
8. If 2.00 mL of 1.00 M HCI is added to the buffer above, what is the resulting pH? X=1.822224507x10
[HA] = 0+0498 [H38] = 080182 [A] = 0.01481.
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