September 29, 2026 at 6:10 pm #9635
Lösung:
With pH = -log(c(H3O+)) following:: c(H3O+) = 10-pH
It is the same.
- 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 - 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 - 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