Homework Answers
Part A)
[propionate] = 0.060M
[acid] = 0.080M
pKa of propionic acid = 4.88
The solution is a buffer and its pH is calculated using Hendersen equation as
pH = pKa + log [ conjugate base]/[acid]
= 4.88+ log 0.060/0.080
= 4.755
Part B
this is a base and conjugate acid buffer
[methylamine] = 0.080 M
[ salt] = 0.11 m and
pKb of methylamine = 4.19
The pOH of this buffer by Hendersen equation is
pH = pKb + log [salt]/[base]
= 4.19 + log 0.11/0.08
= 4.328
and pH = 14-pOH= 14- 4.328 =9.6717
Part C
[acetate] = 0.160M
[acid] = 0.20M
pKa of acetic acid = 4.75
The solution is a buffer and its pH is calculated using Hendersen equation as
pH = pKa + log [ conjugate base]/[acid]
= 4.75+ log 0.160/0.20
= 4.653
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calculate the pH of each solution.
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a buffer contains 0.13 mol of propionic acid and 0.30 mol of
sodium propionate in 1.20 L.
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Calculate the pH of each solution.
Part A) [OH-] = 9.9x10^-7 M
Part B) [OH-] = 3.6x10^-8 M
Part C) [OH-] = 7.2x10^-11 M
Part D) [OH-] = 9.5x10^-2 M
My Delve - Google Dr. < Chapter 14 HW Exercise 14.73 Calculate the pH of each solution Part A (OH) = 9.9x10-7M Express your answer using two decimal places. N AEV O Submit Request Answer Part B [OH - 3.6x10-M Express your answer using two decimal places. 0 AED +...
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