obliczenie pH
: 15 sty 2015, o 22:10
Witam, Czy ktoś mógłby mi pomóc obliczyć pH 0,1 M HCl i 0,1 M CH3COOH?
1) pH=-logCHCl=1
2) [H+]=(KCH3COOH * CCH3COOH)1/2
-log[H+]=pH=1/2(-logKCH3COOH-logCCH3COOH)=1/2(pKCH3COOH-logCCH3COOH)=1/2(pKCH3COOH-logCCH3COOH)=1/2(4.76-(-1))=2.88