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# calculation

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

ans:13.09

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

### 2 Answers

- 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