pH calculations, 1 (SB-47) Calculate the pH of the following solutions if the solvent is always pure water: 1 g Hydrochloric acid (HCl) dissolved in one litre of water 1 kg Hydrochloric acid (HCl) dissolved in a cubic meter of water 1 g NaOH, solved in a 2 1 litre of water 81 mg HBr in 0.1 1 litre of water 1 1 gram of sulphuric acid in 1 m3 Assumption: complete deprotonation of the sulfuric acid.

  Aufgabe: SB-47    

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Calculate the pH of the following solutions if the solvent is always pure water:

1. 1 g Hydrochloric acid (HCl) dissolved in one litre of water
2. 1 kg Hydrochloric acid (HCl) dissolved in a cubic meter of water
3. 1 g NaOH, solved in a 2 1 litre of water
4. 81 mg HBr in 0.1 1 litre of water
5. 1 1 gram of sulphuric acid in 1 m³ Assumption: complete deprotonation of the sulfuric acid.

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Lösung:

Notes 1: HCl is a strong acid (pKs < 0), i.e. it is completely dissolved in water to H₃O⁺ and Cl^–.
Notes 2: pH = -log(c(H₃O⁺)) = -log(n/V)
Notes 3: pH + pOH = 14
Notes 4: HBr is also a strong acid

1. (a) pH = -log1/36.5/1) = 1.56
2. (b) pH = -log1000/36.5/1000) = 1.56
3. c) pOH = -log1/40/2) = 1.90, pH = 14 – pOH = 12.1
4. (d) pH = -log0.081/81/0.1) = 2
5. (e) Sulphuric acid, H₂SO₄ There are two H⁺ Get out of here!
       pH = -log2·1/98/1000) = 4.69

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