Chemistry-titration+ pH

Because it requires very small amount of excess acid to change the pH of a neutral solution (given that there're no complications like buffers).

For example, after complete neutralization, the final volume = 50ml and pH = 7 .
Addition of 0.1ml 1M hydrochloric acid (in excess) will provide 0.0001 mol H(+)
then, pH of resultant solution = -log[H(+)] = -log (0.0001 / (0.050+0.1))
= 2.70

See? even addition of very small amount of excess acid would already change the pH greatly; and the error contributed by that 0.1ml is actually small. Therefore we can safely take the final state (with a drop of excess acid) as the end point.


Of course there're many complications in real life, but the principle is like this.


2013-04-03 12:32:45 補充：
You could have add supplementary to your question.

Actually it's similar to the previous situation.
For example, after methyl orange turns yellow at pH=4.4, the final volume = 50ml.
no. of mole of H(+) in solution = 10^(-4.4) * 0.050 = 1.99*10^-6 mol

2013-04-03 12:35:29 補充：
To neutralize this portion of unreacted H(+) with 0.1M NaOH,
volume = (1.99e-6 / 0.1) * 1000 = 0.02ml
which is just insignificant. Who cares about 0.02ml ?

Again, practically there're complications, but what shown above just to shows you the basic concept.

Thank you very very very much. It helps me a lot :)))