Question

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Explanation:

Step1: Calculate HF converted

The initial concentration of HF is 0.050 M, and at equilibrium, it's 0.030 M. The amount converted is $0.050 - 0.030 = 0.020$ M.

Step2: Stoichiometry for F⁻

From the reaction $\ce{HF + H2O <=> F^- + H3O^+}$, the mole ratio of HF to F⁻ is 1:1. So, $(0.020\ \text{M HF}) \times \frac{1\ \text{mol F}^-}{1\ \text{mol HF}} = 0.020\ \text{M F}^-$.

Step3: Stoichiometry for H₃O⁺

The mole ratio of HF to H₃O⁺ is also 1:1. Thus, $(0.020\ \text{M HF}) \times \frac{1\ \text{mol H}_3\text{O}^+}{1\ \text{mol HF}} = 0.020\ \text{M H}_3\text{O}^+$.

Step4: Calculate \(K_a\)

The formula for \(K_a\) of HF is $K_a = \frac{[\ce{F^-}][\ce{H3O^+}]}{[\ce{HF}]}$. Substituting the equilibrium concentrations: $[\ce{F^-}] = 0.020$ M, $[\ce{H3O^+}] = 0.020$ M, $[\ce{HF}] = 0.030$ M. So, $K_a = \frac{(0.020)(0.020)}{0.030} = \frac{0.0004}{0.030} \approx 0.0133$.

Answer:

1. \(0.020\ \text{M}\)
2. \(1\ \text{mol}\)
3. \(0.020\ \text{M}\)
4. \(0.020\ \text{M}\)
5. \(0.020\ \text{M}\)

\(K_a \approx 0.013\) (or \(\frac{0.020\times0.020}{0.030} \approx 0.013\))