Koliki je pH rastvora...

Submitted by animallover95 on Sat, 04/19/2014 - 19:31
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Koliki je pH rastvora dobijenog rastvaranjem 0.6 molova NaOH i 0.8 molova CH3COOH u određenoj količini vode : Kk(CH3COOH)=1.8x10(-5)... rešenje je 5.22

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Boris - Sat, 04/19/2014 - 19:58

reagovanje NaOH i CH3COOH i nastace CH3COONa(0.6mol) i ostace visak sircetne kiselin(0.2mol), odnosno pufer.

 

c(H+)=Kk*n(CH3COOH)/n(CH3COONa)

c(H+)=6*10-6

pH=5.22

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animallover95 - Sat, 04/19/2014 - 20:07

Hvala puno...!!! :)