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In titration, it is said to run HCL, for example, from a burette into the flask, untill the solution just turns into a persistent orange colour. (suppose two drops of methyl orange indicator is added in a conical flask containing NaOH). At this point, just enough acid has been added to neutralize the alkali.
But according to the colour distribution of methyl orange indicator, orange colour indicates the pH value between 3.1 to 4.4, then how could the above orange solution be neutralized completely to get a solution of NaCl?
- 己式庚辛Lv 78 年前最愛解答
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))
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.
- 8 年前
Thank you very very very much. It helps me a lot :)))