pH Calculations- Answers
For each of the following, calculate [H3O+], [OH-], pH, and pOH.
1.
[H3O+] = 1.56 x 10-3 mol/L
pH = -log[H3O+]
pOH = 14 – pH
[OH-] = antilog(-pOH)
ACIDIC
pH = -log(1.56 x 10-3 mol/L) = 2.807
pOH = 14 – 2.807 = 11.193
[OH-] = antilog(-11.193) = 6.410 x 10-12 mol/L
2.
[H3O+] = 0.125 M
pH = -log[H3O+]
pOH = 14 – pH
[OH-] = antilog(-pOH)
ACIDIC
pH = -log(0.125 mol/L) = 0.903
pOH = 14 – 0.903 = 13.097
[OH-] = antilog(-13.097) = 8.00 x 10-14 mol/L
3.
[OH-] = 4.12 x 10-4 mol/L
pOH = -log[OH-]
pH = 14 – pOH
[H3O+] = antilog(-pH)
BASIC
pOH = -log(4.12 x 10-4 mol/L) = 3.385
pH = 14 – 3.385 = 10.615
[H3O+] = antilog(-10.615) = 2.43 x 10-11 mol/L
4.
[OH-] = 0.0250 M
pOH = -log[OH-]
pH = 14 – pOH
[H3O+] = antilog(-pH)
BASIC
pOH = -log(0.0250 mol/L) = 1.602
pH = 14 – 1.602 = 12.398
[H3O+] = antilog(-12.398) = 4.00 x 10-13 mol/L
5.
pH = 3.45
[H3O+] = antilog(-pH)
pOH = 14 – pH
[OH-] = antilog(-pOH)
ACIDIC
[H3O+] = antilog(-3.45) = 3.6 x 10-4 mol/L
pOH = 14 – 3.45 = 10.55
[OH-] = antilog(-10.6) = 2.8 x 10-11 mol/L
6.
pH = 9.76
[H3O+] = antilog(-pH)
pOH = 14 – pH
[OH-] = antilog(-pOH)
BASIC
[H3O+] = antilog(-9.76) = 1.7 x 10-10 mol/L
pOH = 14 – 9.76 = 4.24
[OH-] = antilog(-4.24) = 5.8 x 10-5 mol/L
7.
pOH = 2.45
[OH-] = antilog(-pOH)
pH = 14 – pOH
[H3O+] = antilog(-pH)
BASIC
[OH-] = antilog(-2.45) = 3.5 x 10-3 mol/L
pH = 14 – 2.45 = 11.55
[H3O+] = antilog(-11.55) = 2.8 x 10-12 mol/L
8.
pOH = 12.5
[OH-] = antilog(-pOH)
pH = 14 – pOH
[H3O+] = antilog(-pH)
ACIDIC
[OH-] = antilog(-12.5) = 3.2 x 10-13 mol/L
pH = 14 – 12.5 = 1.5
[H3O+] = antilog(-1.5) = 0.032 mol/L
L. Farrell – Chemistry 12 – pH Calculations – Answers – Page 1 of 9
Answer the following textbook questions:
14.2
Page 566, questions 12-15
Page 569, questions 17-19
Page 572, questions 20-25
L. Farrell – Chemistry 12 – pH Calculations – Answers – Page 2 of 9
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L. Farrell – Chemistry 12 – pH Calculations – Answers – Page 9 of 9

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