Question

For each of the following solutions, calculate the pH, pOH, [H3O+], and [OH-] at 25⁰C. A...

For each of the following solutions, calculate the pH, pOH, [H3O+], and [OH-] at 25⁰C.

A solution of 0.047 M LiOH

A solution of 0.0011 M HNO3

A solution with a pH of 9.12

A solution of 0.010 M HClO3

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Homework Answers

Answer #1

0.047 M LiOH

LiOH is strong base therefore dissociate completely so [LiOH] = [OH-] = 0.047 M

pOH = -log[OH-] = -log (0.047) = 1.33

pH = 14 - pOH = 14 - 1.33 = 12.67

[H3O+] = 10-pH = 10-12.67 = 2.14 X 10-13M

0.0011 HNO3

HNO3 is strong acid therefore dissociate completely thus [HNO3] = [H3O+] = 0.0011 M

pH = -log[H3O+] = -log(0.001) = 2.96

pOH = 14 - pH = 14 - 2.96 = 11.04

[OH-] = 10-pOH = 10-11.04 = 9.12 X 10-12 M

pH of 9.12

[H3O+] = 10-pH = 10-9.12 = 7.59 X 10-10 M

pOH = 14 - pH = 14 - 9.12 = 4.88

[OH-] = 10-pOH = 10-4.88 = 1.31 X 10-5 M

0.010 HClO3

HClO3 is strong acid therefore dissociate completely thus [HClO3] = [H3O+] = 0.010 M

pH = -log[H3O+] = -log(0.010) = 2.0

pOH = 14 - pH = 14 - 2.0 = 12

[OH-] = 10-pOH = 10-12 = 1 X 10-12 M

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