Strong Acids

Acids and Bases Worksheet

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1. Calculating pH from Concentration

Calculate the pH of the following acids. (Assume all are strong acids).

a) 0.5 mol dm⁻³ of hydrochloric acid.

-log(0.5) = 0.30

b) 0.75 mol dm⁻³ of nitric acid.

-log(0.75) = 0.12

c) 3.0 mol dm⁻³ of hydrochloric acid.

-log(3.0) = -0.48

d) 2 mol dm⁻³ of hydrobromic acid.

-log(2) = -0.30

e) 0.7 mol dm⁻³ of sulfuric acid.

[H⁺] = 0.7 x 2 = 1.4 mol dm⁻³
-log(1.4) = -0.15

f) 1.5 mol dm⁻³ HClO₃.

-log(1.5) = -0.18

g) 0.75 mol dm⁻³ of phosphoric acid.

[H⁺] = 0.75 x 3 = 2.25 mol dm⁻³
-log(2.25) = -0.35

Calculate the concentration of the following acids, given their pH.

a) Hydrochloric acid. pH 2.50

10⁻²·⁵ = 0.0032 mol dm⁻³

b) Hydrobromic acid. pH 0.50

10⁻⁰·⁵ = 0.32 mol dm⁻³

c) Nitric acid. pH 0.20

10⁻⁰·² = 0.63 mol dm⁻³

d) Sulfuric acid. pH -0.40

[H⁺] = 10⁰·⁴ = 2.51 mol dm⁻³
[Acid] = 2.51 / 2 = 1.26 mol dm⁻³

e) Phosphoric acid. pH -0.6

[H⁺] = 10⁰·⁶ = 3.98 mol dm⁻³
[Acid] = 3.98 / 3 = 1.33 mol dm⁻³

Calculate the following.

a) Calculate the pH of 100cm³ of 0.6 mol dm⁻³ HNO₃.

-log(0.6) = 0.22
(Note: Volume does not affect pH)

b) 50cm³ of water is added to 150cm³ of 0.5 mol dm⁻³ HCl. What is the pH?

Total Volume = 200cm³
Dilution factor = 150 / 200 = 0.75
New [H⁺] = 0.5 x 0.75 = 0.375 mol dm⁻³
pH = -log(0.375) = 0.43

c) Calculate the concentration of 200cm³ of sulfuric acid that has a pH of 0.20.

[H⁺] = 10⁻⁰·² = 0.63 mol dm⁻³
[Acid] = 0.63 / 2 = 0.32 mol dm⁻³

d) Calculate the concentration of a 100cm³ solution of HCl that has 75cm³ of water added to it. Original pH was 0.33.

Original [H⁺] = 10⁻⁰·³³ = 0.468 mol dm⁻³
Total Volume = 175cm³
Dilution factor = 100 / 175 = 0.571
New [H⁺] = 0.468 x 0.571 = 0.267 mol dm⁻³

e) Calculate the pH of a solution formed when 30cm³ nitric acid (original pH -0.45) was diluted to 100cm³.

Original [H⁺] = 10⁰·⁴⁵ = 2.82 mol dm⁻³
Dilution factor = 30 / 100 = 0.3
New [H⁺] = 2.82 x 0.3 = 0.846 mol dm⁻³
pH = -log(0.846) = 0.07