[exec]

What is the mass of potassium hydroxide, KOH that must be added to 250.0cm³ of 0.10moldm^-3 hydrofluoric acid, HF in order to obtain a solution with pH of 3.3?
[Relative atomic mass, H = 1; O = 16; K = 39; Ka HF = 6.8 x 10^-4]


Attempt of solution:

The objective is to find "At what concentration of HF would the solution be of pH 3.3?"
Using the relation Kac = [H⁺]², and [H⁺] = 10^-3.3, the desired concentration is c = (10^-3.3)² / 6.8 x 10^-4 = 3.7 x 10^-4 moldm^-3.
Initial concentration = 0.10 moldm^-3, thus the reduction in concentration is 0.10 - 3.7 x 10^-4 = 9.96 x 10^-2 moldm^-3.
The number of mole in such concentration is (9.96 x 10^-2) x 0.25dm³ = 2.49 x 10^-2 mol.
Since KOH --> K+ + OH-, and HF <---> H+ + F-, the amount of substance is 1:1 so that the number of mole of KOH required is also 2.49 x 10^-2 mol.
Thus, the mass of KOH required is (2.49 x 10^-2) x (1+16+39) = 1.39g.

Anyway, I've got the wrong answer, please help me... 

[Borek]

Solution contains both acid and conjugated base. Does it sound a bell?

[exec]

err.... unfortunately no...

HF <---> H+  +  F-     and         KOH ----> K+   +   OH-

Then...?

[enahs]

Look at buffer solutions and Henderson-Hasselbalch equations.

[vhpk]

No, it can't make KF but KHF₂, can't dissolve like that