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Calculate the pH of the following solutions if the solvent is always pure water:
- 1 g hydrochloric acid (HCl) dissolved in a litres water
- 1 kg of hydrochloric acid (HCl) dissolved in one cubic meter of water
- 1 g NaOH, dissolved in 2 litres of water
- 81 mg HBr in 0.1 litres of water
- 1 grams sulphuric acid in 1 m3 water. Assumption: complete deprotonation of the sulphuric acid.
Solution: Note 1: HCl is a strong acid (pKs < 0), i.e. it completely decomposes in water to H3O+ and Cl–.
Note 2: pH = -log(c(H3O+)) = -log(n/V)
Note 3: pH + pOH = 14
Note 4: HBr is also a strong acid- a) pH = -log(1/36.5/1) = 1.56
- b) pH = -log(1000/36.5/1000) = 1.56
- c) pOH = -log(1/40/2) = 1.90, pH = 14 – pOH = 12.1
- d) pH = -log(0.081/81/0.1) = 2
- (e) Sulphuric acid, H2SO4 gives off two H+!
pH = -log(2·1/98/1000) = 4.69
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