Question

If 0.1200g of sodium carbonate is dissolved in 50 mL of water and titrated with 0.1000 M HCl, how many mL of HCl will be required to reach the second endpoint?
CO3^-2 + H^+1 --> HCO3^-1
HCO3^-1,+ H^+1 --> H2CO3

Answer 1

Na2CO3 + 2HCl ==> 2NaCl + H2O + O2
Method A.
Convert 0.1200 g Na2CO3 to moles. moles = grams/molar mass.

Using the coefficients in the balanced equation, convert moles Na2CO3 to mols HCl. That will be 2x moles Na2CO3.

M HCl = moles HCl/L HCl, solve for L HCl and convert to mL.

Method B.
Convert 0.1200 g Na2CO3 to moles.

Using the equation to add the first H to Na2CO3, we have
Na2CO3 + HCl ==> NaCl + NaHCO3

Using the coefficients, convert moles Na2CO4 to moles HCl. In this case moles Na2CO3 are same as moles HCl.

Then M HCl = moles HCl/L HCl and convert to mL.
Finally, since this is the mL to add 1 H, then 2x that will be mL to add both H. Method A and method B will give the same answer.