dodajemy 95 cmJ 0.1 m NaOH
= 2.59
dodajemy 100 cmJ 0.1 m NaOH w roztworze tylko NaCl pH = 7
dodaliśmy 105 cmJ 0.1 m NaOH V = 0.205 dcm1 2 3
NaOH 0.005 dcm3 x0.1m/dcm3NaOH =
= 0.0005 mola
0.0005
pOH - - log = _ i0g 0.00244 =
= -(-2.59) = 2.59
pH= 14-2.59= 11.41