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

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First, we need to calculate the moles of each compound in the solution. moles of NaF = 10.0 g NaF / 41.99 g/mol NaF = 0.238 mol NaF moles of HF = 20.0 g HF / 20.01 g/mol HF = 0.999 mol HF We can assume that all of the NaF will dissociate completely in solution, so we will have 0.238 mol of Na+ ions present. The HF, on the other hand, will partially dissociate according to the reaction: HF + H2O <-> H3O+ + F- The equilibrium constant for this reaction (K_a) is given in the problem as 6.8 x 10^-4. We can set up an ICE table to calculate the concentration of H3O+ ions in solution: I HF H2O <-> H3O+ F- C 0.999 M - <-> x x E 0.999-x - <-> x x Using the equilibrium constant expression for this reaction, we can plug in the known values and solve for x (which is the concentration of H3O+ ions in solution): K_a = [H3O+][F-] / [HF] 6.8 x 10^-4 = x^2 / (0.999-x) After solving this equation (either through the quadratic formula or by making simplifying assumptions and approximations), we find that [H3O+] = 1.6 x 10^-3 M. Finally, we can use the definition of pH to calculate the pH of the solution: pH = -log[H3O+] pH = -log(1.6 x 10^-3) pH = 2.54 Therefore, the answer is B.

If 100 G of NaF and 20