If 0.50 L 0.20 M HF solution is mixed with solution A to make a mixed solution with a total volume of 1.00 L. Kw = 1.00x10-14; Ka(HF) = 6.6x104 at T=298K. (a) If solution A is 0.50 L 0.20 M HNO3, calculate the pH of the mixed solution. Ka(HNO3) = 100 (b) If solution A is 0.50 L 0.20 M NHACI, calculate the pH of the mixed solution. Ko(NH3) = 1.8x105 (c) If solution A is 0.50 L 0.20 M HNO2, calculate the pH of the mixed solution. Ka(HNO2) = 5.1x104 Hint: First list an equation that shows electrical neutrality in the solution. Also, neglect the [H*] contribution from water. Then use [H] and the equilibrium constant(s) to express the anion concentrations in the equation you obtained for electrical neutrality, and you will find there is single unknown (H') in the equation