Part A Exercise 15.50: Problems by Topic - Autoionization of Water and pH pH = 8
ID: 554341 • Letter: P
Question
Part A Exercise 15.50: Problems by Topic - Autoionization of Water and pH pH = 8.53 Express your answer using two significant figures. Enter your answers numerically separated by a comma. Calculate H,O ] and (OH ] for each of the following solutions. Submit My Answers Give Up Incorrect; Try Again; 2 attempts remaining Part B pH 11.20 Express your answer using two significant figures. Enter your answers numerically separated by a comma. Submit My Answers Give Up Part C pH = 2.85 Express your answer using two significant figures. Enter your answers numerically separated by a comma.Explanation / Answer
A)
POH = 14 - pH
= 14 - 8.53
= 5.47
we have below equation to be used:
pH = -log [H3O+]
8.53 = -log [H3O+]
log [H3O+] = -8.53
[H3O+] = 10^(-8.53)
[H3O+] = 3.0*10^-9 M
we have below equation to be used:
pOH = -log [OH-]
5.47 = -log [OH-]
log [OH-] = -5.47
[OH-] = 10^(-5.47)
[OH-] = 3.4*10^-6 M
Answers:
[H3O+] = 3.0*10^-9
[OH-] = 3.4*10^-6
B)
POH = 14 - pH
= 14 - 11.2
= 2.8
we have below equation to be used:
pH = -log [H3O+]
11.2 = -log [H3O+]
log [H3O+] = -11.2
[H3O+] = 10^(-11.2)
[H3O+] = 6.3*10^-12 M
we have below equation to be used:
pOH = -log [OH-]
2.8 = -log [OH-]
log [OH-] = -2.8
[OH-] = 10^(-2.8)
[OH-] = 1.6*10^-3 M
Answers:
[H+] = 6.3*10^-12
[OH-] = 1.6*10^-3
C)
POH = 14 - pH
= 14 - 2.85
= 11.15
we have below equation to be used:
pH = -log [H3O+]
2.85 = -log [H3O+]
log [H3O+] = -2.85
[H3O+] = 10^(-2.85)
[H3O+] = 1.4*10^-3 M
we have below equation to be used:
pOH = -log [OH-]
11.15 = -log [OH-]
log [OH-] = -11.15
[OH-] = 10^(-11.15)
[OH-] = 7.1*10^-12 M
Answers:
[H+] = 1.4*10^-3
[OH-] = 7.1*10^-12
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