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dane V1=V2=V=100cm3= 0,1dm3, pH1=2, pH2=3
szukane pH
n1- liczba moli H+ w I roztworze
n2- liczba moli H+ w II roztworze
[H+] = [n1+n2]/Vc....n1=c1*V.....n2=c2*V.....c1=10(-pH1)=10(-2)M
.........................c2=10(-pH2)=10(-3)M, Vc=0,2dm3
[H+] = [10(-pH1)*V + 10(-pH2)*V]/2V
[H+] = [10(-2)M + 10(-3)M]/2 = 5,5*10(-3)M
pH = -log[H+] = -log5,5*10(-3) = 2,26
Po zmieszaniu roztworów pH wyniosło 2,26 jednostek Sorensena.
Semper in altum.........................pozdrawiam