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Cm = 0,0001mol/dm³ HNO3, H⁺ i NO₃⁻
więc jeżeli stężenie procentowe jonów H⁺ = 10⁻⁴ mol/dm³, to podstawiamy do wzoru, że
pH = -log[H⁺]
pH = -log 10⁻⁴
pH = 4
a wiec
Alfa=CH+/Cw
CH+=Cw*alfa
CH+=1*0,0001=0,0001mola/dm3
PH=-Log(0,0001)
PH=4
PH tego roztworu wynosi 4