Răspuns:
c,so2= 0,1ppm= 0,1 mg/L
M,SO2= 64g/mol
n= 0,1mg/l / 64mg/mmol=0,00156mmol/ l --->
c,m= 1,56x10⁻⁶mol/l
1mol.....................1mol
SO2+ H2O---> H2SO3
1,56x10⁻⁶mol...........x
x= n= 1,56x10⁻⁶mol H2SO3
daca acidul sulfuros este total ionizat, [H3O⁺]=[H2SO3]=1,56X10⁻⁶mol/l
pH= -log[H3O⁺]= 6-log1,56
pH=6-0,19= 5,91--. >mediu slab acid
c= 1ppm=1mg/l
n=1,56x10⁻⁵mol/l
pH= 5-0,19= 4,91
Explicație:
