If a solution of hf (ka=6.8×10−4) has a ph of 2.90, calculate the concentration of hydrofluoric acid.

0.0023

Explanation:

1) Balanced equilibrium equation:

    HF ⇆ H⁺ + F⁻

2) Equilibrium constant of the acid, Ka:

        Ka = [H⁺] [F⁻] / [HF]

[HF] is the concentration searchedFrom the stoichiometry 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:

       

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

The correct answer should be "electrons orbit the nucleus in circular motions exactly as shown on the classic bohr model."

Higher temperature = molecules move around faster. and vice versa.