Question

If a solution of HF (Ka=6.8×10−4) has a pH of 2.90, calculate the concentration of hydrofluoric acid.

Answer 1

Answer: 0.0023

Explanation:

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1) First, you must write the balanced equilibrium equation, which will let you to determine the mole ratios of the different species in solution:

    HF ⇆ H⁺ + F⁻

Therefore, the mole ratio is: 1 HF: 1 H⁺ : 1  F⁻

2) Second, write the equilibrium constant of the acid, Ka, which will permit you to state the relationship of the concentrations in equilibrium:

        Ka = [H⁺] [F⁻] / [HF]

   [HF] is the searched concentration of the hydrofluoric acid

   From the mole ratio of the balanced chemical equation [H⁺] = [F⁻]

       Hence:

       Ka = [H⁺] [H⁺] / [HF] = [H⁺]² / [HF]

3) From the pH value you can calculate [H⁺],  using the definition:

•    pH= - log [H⁺]

•    Substitute: 2.90 = - log [H⁺]

 

 Clear [H⁺] using logarithm properties:

         $[H^+]=10^{-2.90}=0.00126$

4) Now you can substitute [H⁺] and the value of Ka in the equation for Ka:

   Ka = [H⁺]² / [HF] ⇒ [HF] = [H⁺]² / Ka = (0.00126)² / (6.8×10⁻⁴) = 0.00233.

Since the value of Ka has two significant figures, you must report the answer with two significan figures, i.e. 0.0023