Anonymous

# 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

Rating

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

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