This question was created from Experiment 5-Standardization of Solutions.pdf
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This question was created from Experiment 5-Standardization of Solutions.pdf


In class we determined that 0.630 g of oxalic acid dihydrate would neutralize 20.0 mL of 0.50 M NaOH. For trial 1 on the table I recorded C: 0.6357 G D: 6.10 mL E. 29.5 mL F. 23.4 mL G: .0234 L I need help filling in H-K and answering the last two questions. Please show work!


Part 2: Standardization of 0.50 M Sodium Hydroxide Solution
Calculate the mass of oxalic acid dihydrate, H2C204:2H20, that would neutralize 20.0 mL of 0.50
M NaOH solution.
Standardization of Solution (from part 1)
Trial 1
Trial 2
Trial 3
C. Mass of oxalic acid dihydrate (g)
D. Initial volume reading of NaOH (ml)
E. Final volume reading NaOH (ml)
F. Volume of NaOH dispensed (E-D) (ml)
G. Volume of NaOH in liters (F/1000)
H. Moles of NaOH consumed at endpoint (moles)
I. Concentration of NaOH (H/G) (M)
J. Percent difference
K. Average percentage
Calculation of moles of NaOH: (Use the balanced chemical equation and stoichiometry.)
Calculation of actual concentration of NaOH:
Percent difference of theoretical from actual concentration of NaOH:
|theoretical(B) - actual (1)I
x 100% = Percent difference

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