Acetatni pufer

Submitted by Daljinski1234 on Mon, 12/23/2013 - 17:05
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U kom molskom odnosu se nalaze CH3COOH i CH3COONa u acetatnom puferu ciji je pH=5?

K(CH3COOH)=2 x 10-5. Ja sam mislio da odnos slabe kiseline i njene soli u puferu treba da bude 1:1.

ALI,resenje je 1:2.

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milicaradenkovic - Mon, 12/23/2013 - 18:13

CH3COOH i CH3COONa

CH3COOH--------------->CH3COO- + H+

Ka={CH3COO-}*{H+}/{CH3COOH}

pH=5 pa je {H+}=1*10(-5) mol/l

2*10(-5)=n(CH3COO-}/V*1*10(-5}/n(CH3COO-)/V skratis zapreminu i dobijas

n(CH3COO-)/n(CH3COOH)=2*10(-5)/1*10(-5)

n(CH3COO-)/n(CH3COOH)=2

n(CH3COOH):n(CH3COONa)=1:2

n(CH3COOONa)=n(CH3COO-)