Question

A 15 mL sample of hydrofluoric acid, Ka=6.8x10^-4 was titrated with .475 M KOH. The equivalence point was reached after addition of 19.35 mL of base. Determine the molar concentration of the original hydrofluoric acid solution, and find the pH of the solution.

Answer 1

Step 1: Determine the KOH moles quantity:

Using Molarity(M)=mol/vol (L), it is obtained: mol=M*vol, mol =(0,475M)(0,01935 L) = 0,00919 mol KOH

Step 2: Determine the acid concentration:

The reaction is: HF + KOH = KF + H2O, the relation in the stoichiometric coefficients are 1:1, so in the equivalence point we have the same mol quantity of the acid and the KOH.

The concentration of the acid will be the acid mol in the intial volume (15 ml = 0,015 L) that we had in the sample: 0,00919 mol HF / 0,015 L = 0,613 M.

Step 3: pH calculation.

We have to take into account that the acid equilibrium reaction is: HF = H⁺ + F^-

	HF	=	H+	+	F-
Initial conditions	0,613 M
Change	-x		x		x
Equilibrium	0,613-x		x		x

The equilibrium constant expression is: Ka = [H+][F-]/[HF], replacing the obtained values:

6,8x10^-4=x²/0,613-x, the x value is 4,17 x 10^-4 and is the H⁺ concentration.

The pH can be calculated with the expression pH = -log([H⁺]), pH=1,69 :)