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2014-05-03T21:10:25+07:00

CH3COOH (aq) + NaOH (aq) -->CH3COONa + H2O 

(perbandingan koef) 
n CH3COOH= 100ml x 0,1 =10 - (x/40)mmol. Karena asam lemah harus sisa. 
n NaOH = x/40 
n CH3COONa=x/40 

CH3COOH dan CH3COONa membentuk bufer 
maka: 
H+ = Ka x A/G 
10^-5 = 10^-5 x (10-x/40) / (x-40) 
x=200mg <--

Semoga membantu :)
1 5 1
2014-05-03T21:19:39+07:00
PH= H+ = 10^-5

H+ = Ka. n asam lemah/ n anion garam
10^-5 = 10^-5 . 10/n NaOH
10^-5= 10^-6/ n NaOH
n NaOH= 10 mmol

n=M.V
10= 0,1.V
V= 100 ml

maaf kalau salah