✔ 最佳答案
First of all, the solution is prepared by mixing 500 cm3 of ethanoic acid and 500 cm3 of sodium ethanoate, resulting in their initial concentrations being 0.1 M for both.
Then, with reference to the deductions below:
圖片參考：
http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazyequil1.jpg
From this, we can see that the change in concentrations to both species are negligible and hence we may state that both species still stay at a concentration of 0.1M.
When 10 cm3 of 0.1M HCl is added,
Increase in no. of moles of ethanoic acid = 0.001
Decrease in no. of moles of ethanoate ions = 0.001
So,
New no. of moles of ethanoic acid = 0.101
New no. of moles of ethanoate ions = 0.099
Then, with reference to the following equation:
圖片參考：
http://i117.photobucket.com/albums/o61/billy_hywung/Jun07/Crazyequil2.jpg
We can see that the change in pH is given by:
△pH = log (0.099/0.101)
= -0.009
