Reply To:pH-calculations, 2 (SB-48) Calculate The Weight to H3O+ and OH– in … a) … a Litre Water. pH = 6 b) … In the Lake Constance. The pH be 7

Lösung:

With pH = -log(c(H₃O⁺)) following:: c(H₃O⁺) = 10^-pH
It is the same.

1. With pH = -log(c(H₃O⁺)) following::
       c(H₃O⁺) = 10^-pH Responsibility. c(OH^–) = 10^-pOH
       It is the same. Applicable c=n/V=m/M/V and This means that:
       m = c·V·M
2. a) m(H₃O⁺) = c·V·M=10^-6 mol/l · 1.0 l · 19 g/mol = 1.9·10^-5 Grams
       m(OH^–) = 10^-8 mol/l · 1.0 l · 17 g/mol = 1.7·10^-7 Grams
3. b) Volumes Lake Constance, coarse Estimated: l= 50 km, b= 2In other words:, h=300m
       V= 30km³=30·(1000m·1000m·1000m)=10⁹ m³ = 10¹² Litre
       m(H₃O⁺) = c·V·M=10^-7 mol/l · 10¹² l · 0.019 kg/mol = 1900 kg
       m(OH^–) = 10^-7 mol/l · 10¹² l · 0.017 kg/mol = 1700 kg