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?
2NaOH + H2Ox ==> Na2Ox + 2H2O
mols NaOH = M x L = 0.1 x 0.1 = 0.01
mols H2Ox = 0.01 x (2 mols NaOH/1 mol H2Ox) = 0.01 x 2 = 0.02 mols NaOH
Then M H2Ox = mols H2Ox/L H2Ox. You know M and you know mols, solve for L. Convert to mL if desired.
Well, well, looks like we have a classic acid-base titration question here! Don't worry, my friend, I'll make it a little lighter for you.
So, let's do some math! Since oxalic acid is a diprotic acid, it donates two protons. And since NaOH is a monoprotic base, it accepts only one proton. So, we need to take that into account.
We can use the equation: moles of acid = moles of base to figure this out.
First, let's calculate the moles of NaOH. We have 0.1000 M NaOH in 100.00 mL, so that's 0.1000 mol/L x 0.1000 L = 0.0100 moles of NaOH.
Now, since the acid-to-base ratio is 2:1, we need twice as many moles of acid. Therefore, we need 2 x 0.0100 moles = 0.0200 moles of oxalic acid.
Finally, we can calculate the volume of 0.1000 M oxalic acid needed to neutralize this. We already know the moles we need, so let's rearrange the equation to find volume:
moles = molarity x volume
0.0200 moles = 0.1000 M x volume
Solving for volume gives us:
volume = 0.0200 moles / 0.1000 M = 0.200 L or 200 mL
So, you'll need 200 mL of 0.1000 M oxalic acid to completely neutralize 100.00 mL of 0.1000 M NaOH. Voilà!
To calculate the volume of oxalic acid required to neutralize the NaOH, we need to use the concept of stoichiometry.
The stoichiometric equation for the reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH) is as follows:
2H2C2O4 + 2NaOH -> Na2C2O4 + 2H2O
From the balanced equation, we can see that it takes two moles of oxalic acid to react with two moles of sodium hydroxide, producing two moles of water and one mole of sodium oxalate.
Given that the concentration of NaOH is 0.1000 M and the volume is 100.00 mL (convert to liters):
V(NaOH) = 100.00 mL = 0.10000 L
To determine the moles of NaOH, we can use the formula:
moles = concentration (M) x volume (L)
moles(NaOH) = 0.1000 M x 0.1000 L = 0.0100 moles
Since the stoichiometric ratio between NaOH and H2C2O4 is 1:2, we can calculate the moles of oxalic acid required:
moles(H2C2O4) = 0.0100 moles NaOH x 2 = 0.0200 moles
Now we can calculate the volume of 0.1000 M oxalic acid required using the moles and concentration:
V(H2C2O4) = moles / concentration
V(H2C2O4) = 0.0200 moles / 0.1000 M = 0.200 L = 200.00 mL
Therefore, 200.00 mL of 0.1000 M oxalic acid would be required to neutralize 100.00 mL of 0.1000 M NaOH.
To determine the volume of oxalic acid required to completely neutralize the given amount of NaOH, we need to consider the acid-base reaction between oxalic acid (H2C2O4) and sodium hydroxide (NaOH).
The balanced equation for the reaction is:
H2C2O4 + 2NaOH → Na2C2O4 + 2H2O
From the equation, we can see that the stoichiometric ratio between H2C2O4 and NaOH is 1:2. This means that for every 1 mole of NaOH, we need 1/2 mole of H2C2O4.
Given that the concentration of NaOH is 0.1000 M and the volume is 100.00 mL (0.10000 L), we can calculate the number of moles of NaOH using the formula:
moles of NaOH = concentration × volume
moles of NaOH = 0.1000 M × 0.10000 L
moles of NaOH = 0.01000 moles
Since the stoichiometric ratio is 1:2, we need twice as many moles of H2C2O4. Therefore:
moles of H2C2O4 = (1/2) × moles of NaOH
moles of H2C2O4 = (1/2) × 0.01000 moles
moles of H2C2O4 = 0.005000 moles
Now, to calculate the volume of H2C2O4, we need to know the concentration of the oxalic acid solution. The problem states that the concentration is 0.1000 M.
We can rearrange the formula: moles = concentration × volume to solve for volume:
volume of H2C2O4 = moles ÷ concentration
volume of H2C2O4 = 0.005000 moles ÷ 0.1000 M
volume of H2C2O4 = 0.05000 L or 50.00 mL
Therefore, to completely neutralize 100.00 mL of 0.1000 M NaOH, we would need 50.00 mL of 0.1000 M oxalic acid.