pH of Strong Bases

Acids and Bases Worksheet

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1. Calculating pH of Bases

Calculate the pH of the following bases. (Assume K_w = 1.00 × 10^-14)

a) 0.5 mol dm⁻³ sodium hydroxide.

[H⁺] = 1×10⁻¹⁴ / 0.5 = 2×10⁻¹⁴ mol dm⁻³
pH = -log(2×10⁻¹⁴) = 13.70

b) 0.2 mol dm⁻³ potassium hydroxide.

[H⁺] = 1×10⁻¹⁴ / 0.2 = 5×10⁻¹⁴ mol dm⁻³
pH = -log(5×10⁻¹⁴) = 13.30

c) 0.01 mol dm⁻³ barium hydroxide.

[OH⁻] = 0.01 × 2 = 0.02 mol dm⁻³
[H⁺] = 1×10⁻¹⁴ / 0.02 = 5×10⁻¹³ mol dm⁻³
pH = -log(5×10⁻¹³) = 12.30

d) 2 mol dm⁻³ rubidium hydroxide.

[OH⁻] = 2 mol dm⁻³ (Group 1 Base)
[H⁺] = 1×10⁻¹⁴ / 2 = 5×10⁻¹⁵ mol dm⁻³
pH = -log(5×10⁻¹⁵) = 14.30

2. Calculating Concentration from pH

Calculate the concentration of the following bases based on their pH.

a) 100cm³ of sodium hydroxide solution that has a pH of 13.80.

[H⁺] = 10⁻¹³.⁸ = 1.58×10⁻¹⁴
[OH⁻] = 1×10⁻¹⁴ / 1.58×10⁻¹⁴ = 0.63
[NaOH] = 0.63 mol dm⁻³

b) 200cm³ of barium hydroxide solution that has a pH of 12.60.

[H⁺] = 10⁻¹².⁶ = 2.51×10⁻¹³
[OH⁻] = 1×10⁻¹⁴ / 2.51×10⁻¹³ = 0.04
[Ba(OH)₂] = 0.04 / 2 = 0.02 mol dm⁻³

c) 500cm³ of rubidium hydroxide solution that has a pH of 14.48.

[H⁺] = 10⁻¹⁴.⁴⁸ = 3.31×10⁻¹⁵
[OH⁻] = 1×10⁻¹⁴ / 3.31×10⁻¹⁵ = 3.02
[RbOH] = 3.02 mol dm⁻³

3. Complex Calculations

Calculate the following.

a) Calculate the pH of 0.2 mol dm⁻³ strontium hydroxide.

[OH⁻] = 0.2 × 2 = 0.4 mol dm⁻³
[H⁺] = 1×10⁻¹⁴ / 0.4 = 2.5×10⁻¹⁴ mol dm⁻³
pH = -log(2.5×10⁻¹⁴) = 13.60

b) 100cm³ of 2 mol dm⁻³ NaOH is diluted by adding 100cm³ of water. What is the pH?

Total Volume = 200cm³
Dilution factor = 100 / 200 = 0.5
New [OH⁻] = 2 × 0.5 = 1 mol dm⁻³
[H⁺] = 1×10⁻¹⁴ / 1 = 1×10⁻¹⁴ mol dm⁻³
pH = -log(1×10⁻¹⁴) = 14.00

c) Calculate the concentration of a solution of potassium hydroxide that has a pH of 12.80.

[H⁺] = 10⁻¹².⁸ = 1.58×10⁻¹³
[OH⁻] = 1×10⁻¹⁴ / 1.58×10⁻¹³ = 0.0631
[KOH] = 0.0631 mol dm⁻³

d) Calculate the change in pH of 100cm³ 1 mol dm⁻³ KOH when 400cm³ of water is added to it.

Dilution factor = 100 / 500 = 0.2
New [OH⁻] = 1 × 0.2 = 0.2 mol dm⁻³
[H⁺] = 1×10⁻¹⁴ / 0.2 = 5×10⁻¹⁴ mol dm⁻³
pH = -log(5×10⁻¹⁴) = 13.30

e) Calculate the pH of a solution made from 500cm³ of LiOH (pH 14.2) and 300cm³ of water.

Original [H⁺] = 10⁻¹⁴.² = 6.31×10⁻¹⁵
Original [OH⁻] = 1×10⁻¹⁴ / 6.31×10⁻¹⁵ = 1.58 mol dm⁻³
Dilution factor = 500 / 800 = 0.625
New [OH⁻] = 1.58 × 0.625 = 0.991 mol dm⁻³
New [H⁺] = 1×10⁻¹⁴ / 0.991 = 1.01×10⁻¹⁴
New pH = -log(1.01×10⁻¹⁴) = 13.99

f) Calculate the amount of water required to dilute 100cm³ of 0.5mol dm⁻³ NaOH so that it has a pH of 13.30.

Final [H⁺] = 10⁻¹³.³ = 5×10⁻¹⁴
Final [OH⁻] = 1×10⁻¹⁴ / 5×10⁻¹⁴ = 0.1995 mol dm⁻³
Dilution factor required = 0.1995 / 0.5 = 0.399
Total Volume = 100 / 0.399 = 250.63 cm³
Added water = 250.63 – 100 = 150.63 cm³

Mr Cole Chemistry