係咪計左後,將in NaOH 減番佢
將個數-log 佢?
is always a constant
其實有時真係自己推下式-.-
我都想問下d野
20ml 0.1M NaOH + 20ml 0.1 CH3COOH 既pH係幾多
佢畀左Ka=1.76X10^-5
j32738 發表於 22-2-2009 13:01 http://www4.nakuz.com/bbs/images/common/back.gif
As NaOH completely dissocate in water &
no. of mole of NaOH=no. of mole of CH3COOH,
they are in the equivalence point.All CH3COOH change
to CH3COO- ion.Let CH3COOH=HA
A-+ H2O = HA + OH-
At start:0.05 0 0 p.s =0.05,as add same V
At equm:0.05-x x x of NaOH,V x 2,concentration/2
Kb=x^2 /0.05-x
As Kw/Ka=Kb,
1x10^-14 /1.76X10^-5 =x^2 /0.05-x
x=5.3^-6
pH=14 -pOH
=14-(-log)
=8.70//