Introductory to chem course Stoichometry question wrong answer, can anyone fix it

[land_of_ice]

Question:

A 25 mL sample of H2C204 solution required 19.62 mL of 0.341 M NaOH for neutralization. Calculate the molarity of the acid?

The reaction was :
H2C204 + 2NaOH ---------------------> Na2C2O4 + 2H2O

Attempt:
[Which line has the mistake in it and what is it?]

converted 25mL to L
converted 19.62 to L
multiplied 0.01962 by 0.341 m/L of NaOH
and multiplying all of that by 1 mol of H2C204 / for every 2 moles of NaOH
multplied all of that by .025 L to get 0.00323 moles of the H2C204 , they said they wanted the molarity so molarity is moles/liter of solution, you get .00323/0.025L = .129 Molarity of H2C204
THE ANSWER CAME OUT WRONG?? WHAT the heck? Why? Which line has the mistake in it and what is it?

The answer is supposed to be 0.134 right?

 

[DeadFishFactory]

What I do is convert everything to moles (mmoles) in order to figure it out.

19.62 mL * 0.341 M NaOH = 6.69 mmoles OH-

To completely neutralize that, you need an equivalent in H+, so you need 6.69 mmoles H+. However, H2C2O4 produces 2 H+ per dissociation, so you would need half that amount, or 3.35 mmoles H2C2O4. Then divide by its volume.

3.35 mmoles H2C2O4 / 25 mL = 0.134 M H2C2O4 solutionI got 0.134 M. Are you saying you put that and got it wrong? Your answer says that you got 0.129 M. I think the error was in your calculation rather than concept.

 

[Borek]

multplied all of that by .025 L to get 0.00323 moles of the H2C204

If I understand you correctly you have calculated number of moles of oxalic acid and multiplied it by volume to get number of moles of oxalic acid?

And then you divided it back by the same volume?

Strangely, numbers you have listed don't confirm your description. Check why you got 0.00323 and not 0.00334.

--

 

[ktgster]

A 25 mL sample of H2C204 solution required 19.62 mL of 0.341 M NaOH for neutralization. Calculate the molarity of the acid?

H2C204 + 2NaOH ---------------------> Na2C2O4 + 2H2O

easy way of doing stoich is with unit analysis, so in order to that we need everything in Liters and since I don't like dealing with mL/mmol that's just what I'm going to do
0.025 L H2C204
0.01962 L and 0.341 mol/L NaOH

we want to find the molarity of our acid so we want mol/L of H2C204

(0.01962 L/1) * (0.341 mol/1L) * (1 mol H2C2O4/2 mol NaOH) * (1/0.025L) = 0.133 mol/L

basicly you found the mols of your given and multiplied that by your mol ratio (required on top, given on bottom) and then divide that by your volume of the substance you want to find in order to get mol/L

and they way I just showed you, you can do it in one step.

 

[DeeNos]

There are a few mistakes in the attempt at solving the problem. First, the conversion of 25 mL to L is correct, but the conversion of 19.62 mL to L should be 0.01962 L, not 0.1962 L.

Second, when multiplying by 1 mol of H2C204 for every 2 moles of NaOH, the correct ratio should be 1 mol of H2C204 for every 1 mol of NaOH. This is because the balanced equation shows a 1:1 ratio between H2C204 and NaOH.

Lastly, when multiplying by 0.025 L to get moles of H2C204, the correct value should be 0.01962 L, as this is the volume of NaOH used in the neutralization reaction.

Corrected attempt:
Converted 25 mL to L
Converted 19.62 mL to L
Multiplied 0.01962 L by 0.341 mol/L of NaOH
Multiplied by 1 mol of H2C204 for every 1 mol of NaOH
Multiplied by 0.01962 L to get 0.00669 moles of H2C204
Divided by 0.025 L to get 0.2676 Molarity of H2C204

The correct answer is 0.268 Molarity of H2C204. The mistake was in using the wrong ratio of moles between H2C204 and NaOH.