2. 60.0 mL of 0.100 M NH, (K-1.8 X 10-5) is titrated with 0.150 M HCI. Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = 6mmo NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol I 6 C 15 1.5 4.5mm 4.5mmol [N] 70mL kw kb = 110-H =5.56 PH = pKa + log (bass) pKa = log (S56x107) *109 1.8x10-3 =0.0643 pka = 9.26 [NH] 1.5mmol 70ML = 0.0214 PH-9.26+109 60.0643) PH=9.747 For the titration in question 2, calculate the pH at the equivalence point.
2. 60.0 mL of 0.100 M NH, (K-1.8 X 10-5) is titrated with 0.150 M HCI. Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = 6mmo NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol I 6 C 15 1.5 4.5mm 4.5mmol [N] 70mL kw kb = 110-H =5.56 PH = pKa + log (bass) pKa = log (S56x107) *109 1.8x10-3 =0.0643 pka = 9.26 [NH] 1.5mmol 70ML = 0.0214 PH-9.26+109 60.0643) PH=9.747 For the titration in question 2, calculate the pH at the equivalence point.
General Chemistry - Standalone book (MindTap Course List)
11th Edition
ISBN:9781305580343
Author:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Publisher:Steven D. Gammon, Ebbing, Darrell Ebbing, Steven D., Darrell; Gammon, Darrell Ebbing; Steven D. Gammon, Darrell D.; Gammon, Ebbing; Steven D. Gammon; Darrell
Chapter16: Acid-base Equilibria
Section: Chapter Questions
Problem 16.91QP: A 50.0-mL sample of a 0.100 M solution of NaCN is titrated by 0.200 M HCl. Kb for CN is 2.0 105....
Related questions
Question
![2.
60.0 mL of 0.100 M NH, (K-1.8 X 10-5) is titrated with 0.150 M HCI.
Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = 6mmo
NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol
I 6
C 15
1.5
4.5mm
4.5mmol
[N] 70mL
kw
kb
=
110-H =5.56 PH = pKa + log (bass)
pKa = log (S56x107)
*109
1.8x10-3
=0.0643
pka = 9.26
[NH]
1.5mmol
70ML
= 0.0214 PH-9.26+109 60.0643)
PH=9.747](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e1b8531-f821-4d78-bced-c5927dcb7e7e%2Ff153d944-c4f0-4236-9a2b-df6122897645%2F76zwkgy_processed.png&w=3840&q=75)
Transcribed Image Text:2.
60.0 mL of 0.100 M NH, (K-1.8 X 10-5) is titrated with 0.150 M HCI.
Calculate the pH after the addition of 10.0 mL acid. 0.1x60m² = 6mmo
NH3+ HCI NH+CI KA=! 0.15 x 10-15 mmol
I 6
C 15
1.5
4.5mm
4.5mmol
[N] 70mL
kw
kb
=
110-H =5.56 PH = pKa + log (bass)
pKa = log (S56x107)
*109
1.8x10-3
=0.0643
pka = 9.26
[NH]
1.5mmol
70ML
= 0.0214 PH-9.26+109 60.0643)
PH=9.747
Transcribed Image Text:For the titration in question 2, calculate the pH at the equivalence point.
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