Twardość

Tw_og= 250mgCaCO_3/dm³ : 50 =5 mval

zasM = twW = 5,6°t : 2,8°t = 2mval

twNW = tw_og – twW = 3 mval

mvalCaCl₂ = $\frac{40 + 2*35,5}{2}$ = 55,5mg/mval

55,5mg/mval * 3mval/dm³ = 166,5mg/dm³

mvalMg(HCO₃)₂ = $\frac{24 + 2(1 + 12 + 3*16)}{2}$ = 73mg/dm³

73mg/mval * 2 mval/dm³= 146 mg/dm³

Stężenia:

0,05mol_NaOH - 1 dm³

X – 0,2dm³

X=0,01mol_NaOH

NaOH + HNO₃ = NaNO₃ + H₂O

1 mol_NaOH – 1 mol_HNO3

0,01 mol_NaOH – 0,01mol_HNO3

1,05g – 1cm³

100g – x

X=95,24cm³

10g – 95,24cm³

Xg – 1000cm³

X=105g

Mol_HNO3= 1 +14+3*16=63g/mol

1mol – 63g

X mol – 105g

X=1,67mol/ dm³

1,67mol – 1 dm³

0,01 mol – x

X= 0,006dm³= 6ml_HNO3