A 0.15M solution of a weak monoprotic acid HX has a pH 2.69
25 cm3 of it is titrated with 0.25M NaOH
Calculate the pH of the titration solution when 25cm3 of 0.25M NaOH is added
i did it in this way
no.of mole of NaOH remained= 0.25x25/1000-0.15x25/1000
but it's wrong.
could you please tell me what is the problem
- Po HungLv 61 decade agoFavorite Answer
You assumed thatthe monoprotic acid is completely dissociated to give the hydrogen ions. However, the pH value shows that it is not completely dissociated.
pH = -(Log[H ]), so 2.69 = -(log [H ]) and [H ] = 2.04x10^-3.
Since there is 25ml of acid, the mole of H is [H ]x25/1000 = 5.1x10^-5.
The amount of OH- in 25ml of 0.25M NaOH is 0.25 x 25/1000 = 6.25 x10^-3
After titration, OH- - H = 6.25x10^-3 - 5.1 x 10^-5 = 6.199x 10^-3
After titration, the final volume is 50 ml. So, the [OH-] is 6.199x 10^-3 x1000/50 = 0.124M.
pH = 14 (log[OH-]) = 14-0.9065 = 13.09.
I hpe you understand what I calculated here.
2008-05-20 13:52:41 補充：
Uncle Michael is correct. I forgot the presence of strong alkali will consequently react all the acid.
The pH should be 12.7 as Uncle Michael showed.Source(s): me
- Uncle MichaelLv 71 decade ago
The answer is NOT 13.09.
Also, answer 001 is incorrect. [H3O^+] in the HX solution is 2.04 x 10^-3 M. When the [H3O^+] is removed by NaOH added, the HX molecules will dissociated to give more [H3O^+], until all HX molecules are dissociated.
2008-05-13 17:09:13 補充：
No.of mole of NaOH remained= 0.25x25/1000-0.15x25/1000 = 0.0025 mol
final volume of NaOH = 25 + 25 = 50 cm^3 = 0.05 dm^5
The hydrolysis of X^- is negligible.
[OH^-] = mol/V = 0.0025/0.05 = 0.05 M
pH = 14 - pOH = 14 -(-log0.05)) = 12.7