25.0 mL of a hydrofluoric acid solution of unknown concentration is titrated with 0.200 M NaOH. After 20.0 mL of the base solution has been added, the pH in the titration flask is 3.00. What was the concentration of the original hydrofluoric acid solution. (Ka (HF) = 7.1 * 10-4)

[HF]₀ = 0.125M

Explanation:

NaOH + HF = > NaF + H₂O

Adding 20ml of 0.200M NaOH into 25ml of HF solution neutralizes 0.004 mole of HF leaving 0.004 mole NaF in 0.045L with 0.001M H⁺ at pH = 3. This is 0.089M NaF and 0.001M HF remaining.

=> 45ml of solution with pH = 3 and contains 0.089M NaF from titration becomes a common ion problem.

HF ⇄ H⁺ + F⁻

C (eq) [HF] 10⁻³M 0.089M (< = soln after adding 20ml 0.200M NaOH)

Ka = [H⁺][F⁻]/[HF]₀ = > [HF]₀ = [H⁺][F⁻]/Ka

[HF]₀ = (0.001) (0.089) / (7.1 x 10⁻⁴) M = 0.125M