Solving Malonic Acid Neutralization: 30mL sample, 35mL 2M NaOH

[DarylMBCP]

Hey guys, I'm having a problem with this question;

Malonic Acid, C₃H₄O₄, is a diprotic acid. How many grams of it are in a 30mL sample that requires 35mL of 2M NaOH for complete neutralization?

Since Malonic Acid is diprotic, I think the chemical equation is;
C₃H₄O₄ + 2NaOH --> Na₂C₃H₂O₄ + 2H₂O .

I knw that if the equation is true, the stoichiometric ratio of NaOH : C₃H₄O₄ is 2 : 1 so the number of moles of Malonic Acid is 0.5 x (2 x (35 / 1000)mol = 0.035mol .

Hence, the mass of Malonic Acid required would be 0.035mol x molar mass. However, how did the 30mL fit in? Thanks.

 

[symbolipoint]

Try to think in problem solving steps. First, how many MOLES of the acid are present in the titrated solution? Now use the formula weight of the malonic acid to convert the moles to grams.

 

[DarylMBCP]

First, how many MOLES of the acid are present in the titrated solution?

I found out alrdy that the number of moles of Malonic Acid is 0.5 x (2 x (35 / 1000)mol = 0.035mol .

Now use the formula weight of the malonic acid to convert the moles to grams.

Yes, the mass of Malonic Acid required would be 0.035mol x molar mass or formula weight. However, I am not sure how I'm supposed to incorporate the 30mL sample part in the solution.

Thanks for the helping though.

 

[symbolipoint]

The volume of solution in which your sample is dissolved is not important. You were interested in the grams (the mass) of the malonic acid present.

I am not sure how I'm supposed to incorporate the 30mL sample part in the solution.

If you were interested in the concentration of malonic acid, then you would have a little more calculation to handle.

 

[DarylMBCP]

Oh, ok; so I don't hve to include the 30mL part in my calculations, right?

 

[Borek]

Yes, you can ignore it. This question is equivalent to:

Malonic Acid, C3H4O4, is a diprotic acid. How many grams of it are in a sample that requires 35mL of 2M NaOH for complete neutralization?

 

[DarylMBCP]

Ok then, I get it. Thanks for the help.