[CHEM]pKa

j32738

係咪計左[H3O+]後,將[OH-]in NaOH 減番佢
將個數-log 佢?

Zend

[H3O+][OH-] is always a constant


其實有時真係自己推下式-.-

woodychung

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

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 [A-]=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[5.3^-6])
   =8.70//