Unknown Solid Acid

Purpose:
The purpose of this lab is to titrate an unknown solid acid (KH2PO4) with a standardized sodium hydroxide solution. After recording and plotting the data, the acid’s equivalence point will be recorded once the color changes. Using the equivalence point, the halfway point will be calculated, which is used to determine the acid’s equilibrium constant. The acid’s calculated equilibrium constant will be compared with the acid’s established pKa value. Eventually using the NaOH and the acid’s consumed moles, the equivalent mass will be determined.
Procedure:
Part 2: Obtain 45mL of NaOH, and then weigh 0.3-0.4g of the unknown acid (KH2PO4). Dissolve the acid into 20.00mL water. Record the buret readings, and slowly titrate the NaOH into

There is one mole of OH- in the solution since NaOH goes to Na+ and OH-.
Trial 1: 25.65mL NaOH x 0.100mol/1000mL = 2.57 x 10-3 mol NaOH = 2.57 x 10-3 mol HA = 2.57 x 10-3 mol H+. The equivalent mass is 0.356g Acid / 2.57 x 10-3 mol H+ = 139g/mol H+
Trial 2: 49.57mL NaOH x 0.100 mol / 1000 mL = 4.96 x 10-3 mol NaOH = 4.96 x 10-3 mol HA = 4.96 x 10-3 mol H+. The equivalent mass is 0.644g Acid / 4.96 x 10-3 mol H+ = 130.g/mol H+
Average = (139g Acid / 1 mol H+) + (130.g Acid / 1 mol H+) / 2 = 135g/mol H+. The average equivalent mass for the acid is 135g/mol H+.
3. The answer obtained in Question #2 is the equivalent mass of the acid rather than the molar mass because the acid could be polyprotic, which would mean the equivalent mass is different from the molar mass since it is depending on moles of H+ per molecule, and there could be multiple moles of H+ ions in one mole of a molecule.
4. The KHP and the acid samples must be dried, because there would still be extra water which would skew the molarity. The calculated molarity of the NaOH would be lower because there would be extra volume in the solution, but still the same amount of moles of NaOH, so the molarity would be less, and thus, will require more titrant in order to