What volume of 0.1000 M oxalic acid (a diprotic acid) would be required to completely neutralize 100.00 mL of 0.1000 M NaOH?
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To find the volume of oxalic acid required to neutralize NaOH, we can use the concept of stoichiometry. In this case, oxalic acid (H2C2O4) is a diprotic acid, meaning each molecule of oxalic acid can donate two protons (H+) in a reaction with a base like NaOH.
The balanced chemical equation for the reaction between oxalic acid and NaOH is as follows:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the balanced equation, we see that for every 1 mole of oxalic acid (H2C2O4), we need 2 moles of NaOH to neutralize it and produce water.
First, we need to determine the number of moles of NaOH present in the given volume (100.00 mL) and concentration (0.1000 M):
Moles of NaOH = Volume (L) × Concentration (mol/L)
= 0.10000 L × 0.1000 mol/L
= 0.01000 mol
Since the stoichiometry ratio is 1:2 between H2C2O4 and NaOH, we can conclude that we need twice the number of moles of NaOH to neutralize H2C2O4.
Therefore, the number of moles of H2C2O4 required is:
Moles of H2C2O4 = 2 × Moles of NaOH
= 2 × 0.01000 mol
= 0.02000 mol
Now we need to find the volume of 0.1000 M H2C2O4 required. We can use the following equation:
Moles = Volume (L) × Concentration (mol/L)
Since we know the number of moles of H2C2O4 required, we can rearrange the equation:
Volume (L) = Moles / Concentration (mol/L)
= 0.02000 mol / 0.1000 mol/L
= 0.2000 L
Finally, we convert the volume from liters to milliliters:
Volume (mL) = Volume (L) × 1000
= 0.2000 L × 1000
= 200.0 mL
Therefore, you would require 200.0 mL of 0.1000 M oxalic acid to completely neutralize 100.00 mL of 0.1000 M NaOH.