Stała autodysocjacji wody KH2O w temperaturze 50°C wynosi 1,067x10-15.Oblicz pH wody w tej temperaturze. odp. 6,61
[oh-]=√k
[oh-} =3,27*10^-8
poh= -log [oh-]
poh = 7,49
ph=14-7,49
ph=6,61
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[oh-]=√k
[oh-} =3,27*10^-8
poh= -log [oh-]
poh = 7,49
ph=14-7,49
ph=6,61