Keilyn Mercado
Chem 103
Acid-Base Lab
Trial
#
Mass of
KHP
titrated
Moles of KHP
titrated
Volume of
NaOH dispensed
at the end point
Molarity of NaOH,
calculated (mol/L)
0.480 g
0.00235 moles
21.89 mL
0.10735 mol/L
0.485 g
0.00237 moles
21.89 mL
0.10826 mol/L
0.491 g
0.00240 moles
22.28 mL
0.10771 mol/L
0.497 g
0.00243 moles
22.77 mL
0.10671 mol/L
0.503 g
0.00246 moles
22.79 mL
0.10794 mol/L
0.513 g
0.00251 moles
0.10689 mol/L
23.48 mL
Average
0.10748 mol/L
Molarity:
1
2
3
4
5
6
Moles of KHP in trial 1
N= 0.480g/204.22 g/mol = 0.00235 moles.
Molarity of Sodium Hydroxide solution in trial 1:
0.00235 mol/ 0.02189 L = 0.10735 mol/L
Part II: answer the following three questions:
1. What is Potassium Hydrogen Phthalate and why is it used for standardization of an
NaOH solution?
o Potassium Hydrogen Phthalate (KHP) is an acidic salt compound. This compound is a
white powder that has colorless crystals and solutions. KHP is commonly used in
acid-base titrations because it not only standardizes solutions of bases, it has high
reactivity and it is less likely to absorb moisture from the air.
2. In a similar experiment to Part I, a solution of calcium hydroxide of unknown
concentration is standardized against potassium hydrogen phthalate (KHP). From the data
below, calculate the molarity of Ca(OH)2 solution. The balanced reaction is: Ca(OH)2 +
2KHC8H4O4  CaC8H4O4 + K2C8H4O4 + 2H2O. Note the 1:2 mole to mole ratio of
calcium hydroxide to KHP.
o Molarity of Ca(OH)2 = 0.0847 M
o Mass of KHP consumed at titration end point: 0.914 g
- 0.0022377324 moles
o Ca(OH)2 titrated to reach endpoint: 26.42 ml
- 0.02642 L
3. Determine how many liters of the standardized Ca(OH)2 in question 2 will be needed to
neutralize an aqueous solution containing 0.1453 g of oxalic acid dihydrate (126.07
g/mol). The balanced reaction is: H2C2O4 + Ca(OH)2 → CaC2O4 + 2 H2O
- 0.0136 L