If 10.0 g of NaF and 20.0 g of HF are dissolved in water to make one liter of solution, what will the pH be? For HF, K_a = 6.8 × 10⁻⁴. A)7.13 B)2.54 C)1.57 D)3.17 E)4.86

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First, we need to write the balanced chemical equation for the reaction between NaF and HF: HF + NaF → NaH + F2 This reaction produces NaH which will react with water to produce Na+ and OH- ions. The HF will also partially dissociate into its ions. Therefore, the solution will contain Na+, F-, H+, and OH- ions. To calculate pH, we need to first find the concentration of H+ ions. This can be done by finding the moles of HF that dissociate (since HF is a weak acid, it only partially dissociates): HF + H2O ⇌ H3O+ + F- The equilibrium constant for this reaction is given by: K_a = 6.8 × 10⁻⁴ Let x be the concentration of H3O+ ions produced by the dissociation of HF. Then, the equilibrium concentrations of HF, H3O+, and F- ions will be (20 - x), x, and (10 + x), respectively. Using the equilibrium constant expression, we get: 6.8 × 10⁻⁴ = (x)(10 + x) / (20 - x) Solving for x, we get: x = 1.22 × 10S1U1P−S1S1P1 M This is the concentration of H3O+ ions in solution. To find the pH, we take the negative logarithm: pH = -log[H3O+] pH = 2.54 Therefore, the answer is choice B.

If 100 G of NaF and 20