obliczenie pH

[magda603]

Witam, Czy ktoś mógłby mi pomóc obliczyć pH 0,1 M HCl i 0,1 M CH3COOH?

[pesel]

1) pH=-logC_HCl=1

2) [H⁺]=(K_CH3COOH * C_CH3COOH)^1/2

-log[H⁺]=pH=1/2(-logK_CH3COOH-logC_CH3COOH)=1/2(pK_CH3COOH-logC_CH3COOH)=1/2(pK_CH3COOH-logC_CH3COOH)=1/2(4.76-(-1))=2.88