4. The titration of 25.0 ml of 0.100 M HCl is made with 0.100 M of NaOH as the titrant. pH=-log CH] ア- PH= 3-l0g5 PM=2.60 a) What is the original pH before adding the base? [H+) - 2.5x10-3 x 0.1 = 2.5X10-4 M 2.5 x 10 3 pM=-log 2.5K(03) b) What is the pH after adding 5.00 ml of the base? 2.0 [4+]= = O.066 M pH= - log (G066) PH= 1.18 CH+]- N,V, ーNV2 Vi+VL (2.5 x0.1) - (5x G1) 2ら+5 c) What is the pH at the equivalence point? At eguivalen ce point, neutrilization tahes place hen PM =7 3ラX 0.1 =Vx 0,1 V 25 mL d) What is the pH at half the equivalence point? At half eguvalence point, only half moles of HCI Pet out in t + of mules of acia lef :- Cuncentra ton of ucl= 1.25 o3 P4 = - log ] 25xi0-3 =(125 N0-3 %3D O0333 (25 +125) x3 Pu=1.48 PH -log (0.0333) e) What is the pH affer adding 30.00 ml of the base? f) Sketch the titration curve.
4. The titration of 25.0 ml of 0.100 M HCl is made with 0.100 M of NaOH as the titrant. pH=-log CH] ア- PH= 3-l0g5 PM=2.60 a) What is the original pH before adding the base? [H+) - 2.5x10-3 x 0.1 = 2.5X10-4 M 2.5 x 10 3 pM=-log 2.5K(03) b) What is the pH after adding 5.00 ml of the base? 2.0 [4+]= = O.066 M pH= - log (G066) PH= 1.18 CH+]- N,V, ーNV2 Vi+VL (2.5 x0.1) - (5x G1) 2ら+5 c) What is the pH at the equivalence point? At eguivalen ce point, neutrilization tahes place hen PM =7 3ラX 0.1 =Vx 0,1 V 25 mL d) What is the pH at half the equivalence point? At half eguvalence point, only half moles of HCI Pet out in t + of mules of acia lef :- Cuncentra ton of ucl= 1.25 o3 P4 = - log ] 25xi0-3 =(125 N0-3 %3D O0333 (25 +125) x3 Pu=1.48 PH -log (0.0333) e) What is the pH affer adding 30.00 ml of the base? f) Sketch the titration curve.
Chemistry: Principles and Reactions
8th Edition
ISBN:9781305079373
Author:William L. Masterton, Cecile N. Hurley
Publisher:William L. Masterton, Cecile N. Hurley
Chapter14: Equilibria In Acid-base Solutions
Section: Chapter Questions
Problem 74QAP: Fifty cm3 of 1.000 M nitrous acid is titrated with 0.850 M NaOH. What is the pH of the solution (a)...
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![4. The titration of 25.0 ml of 0.100 M HCI is made with 0.100 M of NAOH as the titrant.
a) What is the original pH before adding the base?
[H +) - 2.5x(0- x 0.1
=2.5X10-4 M
2.5 x 103
pM=-10g 25KCO-3)
-PH= 3-10g25
PM=2.60
b) What is the pH after adding 5.00 ml of the base?
2.0
- 0. O66 M
pH= - log (G066)
PH=1.18
CH+]- N,V ーNV2
30
Vi+VZ
(2.5 x0.1)-(5x Q1)
2ら+5
c) What is the pH at the equivalence point?
At
eguivalen ce point, neutrilization takes place chen PH =7
3ラX 0.1 -Vx 0.1
V 25 mL
d) What is the pH at half the equivalence point?
At half eguivalence point, only half mules of HCI Pet out in 4
2.5xi0-3
# of moles of acid left
:- (uncentra tin of ucl= 1.25o3
=1.25 O-3
%3D
O0333
PH=-1og(0.0333)
e) What is the pH affer adding 30.00 ml of the base?
PH=- log H+]
(25 +125) x103
pu=1.48
f) Sketch the titration curve.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F71f64add-b02d-4609-b711-53a4d45897f0%2Ff6c294aa-51b1-46fa-ba8b-30342b3f99a1%2Fwxl404l_processed.jpeg&w=3840&q=75)
Transcribed Image Text:4. The titration of 25.0 ml of 0.100 M HCI is made with 0.100 M of NAOH as the titrant.
a) What is the original pH before adding the base?
[H +) - 2.5x(0- x 0.1
=2.5X10-4 M
2.5 x 103
pM=-10g 25KCO-3)
-PH= 3-10g25
PM=2.60
b) What is the pH after adding 5.00 ml of the base?
2.0
- 0. O66 M
pH= - log (G066)
PH=1.18
CH+]- N,V ーNV2
30
Vi+VZ
(2.5 x0.1)-(5x Q1)
2ら+5
c) What is the pH at the equivalence point?
At
eguivalen ce point, neutrilization takes place chen PH =7
3ラX 0.1 -Vx 0.1
V 25 mL
d) What is the pH at half the equivalence point?
At half eguivalence point, only half mules of HCI Pet out in 4
2.5xi0-3
# of moles of acid left
:- (uncentra tin of ucl= 1.25o3
=1.25 O-3
%3D
O0333
PH=-1og(0.0333)
e) What is the pH affer adding 30.00 ml of the base?
PH=- log H+]
(25 +125) x103
pu=1.48
f) Sketch the titration curve.
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