Question

**Problem 7C should have KOH listed as the solution, not NaOH

7. In a different titration, 50.0 mL of 0.150 M HF (pKa = 3.10) was titrated with a solution of 0.250 M KOH. A. (3 pts) The following figure represents solutions at various stages of the titration. (The K ions and water molecules have been omitted for clarity.) Which drawing corresponds to the region of the titration after addition of KOH but before the equivalence point? = HF = F = OH (i) (ii) (iii) (iv) B. (3 pts) At the equivalence point, the pH of the solution being titrated should have a pH of (circle one): greater than 7 less than 7 equal to 7 C. (10 pts) Write a balanced net ionic equation for the reaction that occurs when the NaOH solution was added to the HF solution, including phases of reactants and products. Then determine the pH of the solution after 40.0 mL of the KOH solution has been added.

Please circle the correct answer and show all work for 7c. Thanks

Answer 1

& point, contains solution Before the equivalence He and Fo. figure (i) (Ans). is and strong 6 kr base. salt of weak acid pH770 So than 7 (Ans) for greater ② equation: - Balanced net icnic HA (ar) & OH(aq) - F- (aq) + Hole). * 40 ml of 0.250 M KOH = 0.040 2 x 0.250 ondela - 0.01 mole Mole of HF = 0.050 t X 0.150 mole / 7.5x10-3 mole Excess KOH = (0.01 - 715x103) = 2.5x10 3 mole Total volume = (50 of 40) = 9o me = o.ogoL 2.5x10 3 mole (kot excess = 0.090 L = 0.0278 M pot = -log (0.0278) = 1.56 pH = 14-1.56 – 12.44 TpH = 12.497