Hi I have this problem I have no idea where or what to do.

Question

25.00 mL of 0.0500 M Na2CO3 was titrated with HCl(aq) using phenolphthalein indicator, the 'end-point' (color change from pink to colorless) was reached when 10.50 mL of the acid was added. When methyl orange was used, the end-point (pink to orange) was reached when 21.00 mL of the acid was added. Account for the difference in titer volumes, and calculate the molarity of the acid.

Answer 1

There IS no difference in the volumes. It takes 10.50 mL of the acid to add the first H as follows:
Na2CO3 + HCl ==> NaHCO3 + NaCl

Then it takes another 10.50 mL (21.00-10.50 = 10.50 mL) to add the second hydrogen.
NaHCO3 + HCl ==> NaCl + H2CO3 (=>H2O + CO2).

(The problem did NOT say that ANOTHER 21.00 mL was added for a total volume of 21.00 + 10.50 = 31.50. If that is the case, that's a completely different problem.)
[Another note: The question about the titer could mean that you are to explain why it took twice as much for the second end point as the first end point. Of course the answer is that the first H is being added with the first 10.50 mL and the second H is being added in the next 10.50 mL and it SHOULD take twice as much provided there is no NaHCO3 impurity in the Na2CO3 sample).
As for the M of the acid,
M x L = moles.
You know how many moles were in the Na2CO3 (M x L) and you use moles Na2CO3 = M x L to calculate M HCl.

Answer 2

1.53molar