What determines the strength of a solute as an electrolyte?
1 answer:
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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:
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