Question

Volumetric Determination of a Sulfuric Acid Solution
Initial concentration: assumed 6M H2SO4
	Trial 1	Trial 2
Molarity of NaOH (aq)	0.2053 M	0.2053 M
Initial burette reading, mL	18.50 mL	28.30 mL
Final burette reading, mL	29.50 mL	39.10 mL
a) Moles of NaOH added	2.258 moles	2.217 moles
b) Moles of H2SO4 titrated	?	?

Trial 1:
a)
Final Volume - Initial Volume = Volume
29.50 mL - 18.50 mL = 11.00 mL

Moles = molarity x volume
Moles = 0.2053 M x 11.00 mL
Moles = 2.2583 moles
Moles = 2.258 moles

b)
I don’t know how to calculate this. Please help with Trial 1 b) so I can do Trial 2 b) on my own.

Trial 2:
a)
Volume = 10.80 mL

Moles = 0.2053 M x 10.80 mL
Moles = 2.21724 moles

Answer 1

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