Chapter10: Potentiometry And Redox Titrations
Section: Chapter Questions
Problem 7P
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Correct the answer to the questions. SHOW ALL WORK IN HAND-WRITING:
![What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the
solution above?
Given:
VHCI =1.0 mL
CHC] = 0.10
Concentration of Sulution Albove
- let HA=Acid->HA +
THAT = (25.0mL) (0.10o) – t20.0 mL0.10ml)
(25.0+20.0) mL
[HAJ = 0.011| nal
[A-] =
Of M.
(20-0ML5(0.10) I
(25.0+ 20.0) mL
mol
[A-] = 0.0444 l or mol
[A]
20.0 mL (0.0444 )
(20.0+ 0.1)m
1.0mL (0.101)
L(20.0+1.0) mL
pH =6.4478
Answer:
94](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbfefd57-6cd5-45bf-b890-c343c2fceb41%2F6999297a-a033-4a4f-a239-c8b387209a2d%2Fyetjn3_processed.jpeg&w=3840&q=75)
Transcribed Image Text:What is the pH after the addition of 1.0 mL of 0.10 M HCl to 20 mL of the
solution above?
Given:
VHCI =1.0 mL
CHC] = 0.10
Concentration of Sulution Albove
- let HA=Acid->HA +
THAT = (25.0mL) (0.10o) – t20.0 mL0.10ml)
(25.0+20.0) mL
[HAJ = 0.011| nal
[A-] =
Of M.
(20-0ML5(0.10) I
(25.0+ 20.0) mL
mol
[A-] = 0.0444 l or mol
[A]
20.0 mL (0.0444 )
(20.0+ 0.1)m
1.0mL (0.101)
L(20.0+1.0) mL
pH =6.4478
Answer:
94
![Calculate the amount of solid NHẠC1 to add to 60.0 mL of 0.100 M ammonia to
form a mixture that has the desired pH. Assume no volume change in the solution.
Hint: Use the same buffer ratio that you used in (a).
pH = -log ka + log EVA4, - wknowm
LNH,]
INHT]-> Uu known
[(0.100)]
[x]
1.0=-109(5.81016)+ log
mel
X = 0.1724 = [NH4C1]
Since volune of solution= Constaut:
1724
IL
%3D
mol NHYCI =0.01o34 mol
Molar
Mass NHYCI = 0.01034 (33. 4919) Mass of NtyCI
mol
Mass NHyCl = 0.55346
close
Volume of NH3
Grams of NH4CI
60.00 mL
Di5534.
97](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ffbfefd57-6cd5-45bf-b890-c343c2fceb41%2F6999297a-a033-4a4f-a239-c8b387209a2d%2Fh4tfng4_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Calculate the amount of solid NHẠC1 to add to 60.0 mL of 0.100 M ammonia to
form a mixture that has the desired pH. Assume no volume change in the solution.
Hint: Use the same buffer ratio that you used in (a).
pH = -log ka + log EVA4, - wknowm
LNH,]
INHT]-> Uu known
[(0.100)]
[x]
1.0=-109(5.81016)+ log
mel
X = 0.1724 = [NH4C1]
Since volune of solution= Constaut:
1724
IL
%3D
mol NHYCI =0.01o34 mol
Molar
Mass NHYCI = 0.01034 (33. 4919) Mass of NtyCI
mol
Mass NHyCl = 0.55346
close
Volume of NH3
Grams of NH4CI
60.00 mL
Di5534.
97
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