Question

A 25.00 mL sample of nitric acid requires 19.63 mL of 0.1103 M NaOH to reach the end point of the titration. What is the molarity of the nitric acid solution?

0.536 grams of KHP were added to 100.0mL of water. What is the molarity of the KHP solution? (Do not type units with your answer.)

Following the procedure for today's lab, a 0.0206 M KHP solution requires 24.59 mL of NaOH solution to titrate it.   What is the molarity of the NaOH(aq) solution? (Do not give units with your answer.)

Answer 1

en LTN HNO3 + NAOH Moles of HNO3 Nanog + H2O moles of Naoh 1 = Molanity * Volume (HN) = molarity x volume (NaOH) 19.63 mL Molarity x 25.00 mL = 0.1103 M * → Molarity of HNO3 = 0.08661 M Molar mass of KHP COOH mas of kn (elo) = 204.12 sl mol = 204.22 g/mol -Cook mass → Moles of KHP = 0.536 204.22 g g/mol Molon mass = 2.62 x 10-3 mol Molanity of KHP = moles of kup Volume of solution in L 2.62 x 103 mol 100.0 nył x 10 3 L me = 0.0262 mol/l Answer : 10.0262)

3) KHP + Nooh - Nake + ho moles of KHP = moles of NaOH or Molarity X volume (KHP) = molarity ☆ volume l NaOH) 206 mx volume (RAP) = Molarity (NaOH) X 24.59 mL
In the last part it wasn't clear in the question whether the volume of KHP is 100.0 mL or some other value(it has a reference to lab's procedure which is missing here)...you can put the volume...the next step will be your answer.