In a titration experiment, 45.7 mL of 0.500 M H2SO4 was required to neutralize a 20.0 mL sample of NaOH. Determine the concentration of the NaOH solution.
2007-07-09 04:48:38 · 4 answers · asked by fabulous101 2 in Science & Mathematics ➔ Chemistry
The H2SO4 solution is 0.5 Molar, but 1.0 Normal (because the each mole of acid contributes 2 moles of H+).
Therefore 45.7 mL of 1N acid neutralised 20 mL of NaOH.
Therefore, the NaOH was 45.7/20 Normal =2.285N
Because the NaOH has one OH-, the Molarity is the same as the normality
Hence the NaOH is 2.285M
2007-07-09 05:07:43 · answer #1 · answered by AndrewG 7 · 1⤊ 0⤋
since this reaction involves a mono protic acid, you can assume for every one part of NaOH is consumed by one part of H2S04
First make everything into liters. so we have .0457L of .5 M H2s04 consumed .02 L of NaOH
Solve for the numbers of moles of H2S04, since the number of moles of H2S04 is the same as NaOH
.5M H2S04 * .0457L = .00285 Mol of H2S04
.02285 Mols of H2SO4 were used.
now divide that by the number of Liters of NaOH
.02285/.2 = 1.1425 M NaOH
hope this has helped
2007-07-09 12:03:13 · answer #2 · answered by nathan 6 · 0⤊ 1⤋
titration formula=
molarity x volume = molarity x volume
.5 M H2SO4 x 45.7ml = (variable) x 20ml
22.85= 20 x (variable)
22.85/20 = 1.1425
the molarity of NaOH = 1.1425M
2007-07-09 12:12:22 · answer #3 · answered by Footballfan2345 2 · 0⤊ 2⤋
The reaction is:
H2SO4 + 2NaOH >> Na2SO4 + 2H2O
(0.500moles/L)*0.0457L= 2.29x10-2moles H2SO4
Now considering we have a 2 to 1 ratio we have
2.29x10-2 moles H2SO4*(2/1)=4.57x10-2 moles NaOH
therefore:
4.57x10-2moles/0.020L=2.3M
2007-07-09 20:19:18 · answer #4 · answered by scott k 4 · 0⤊ 0⤋