What volume in millilitres of 0.06176 M HCl is required to titrate 50.00 mL of solution containing 0.02159 M NaHCO3 and 0.0245 M Na2CO3 to a methyl red end point?

Answer 1

I use the titration formula e_AM_AV_A = e_BM_BV_B I'm not sure why this isn't used in books for acids or bases that have multiple protons or hydroxides to donate. A is for acid, B is for base. The e_A stands for the number of H⁺ available in the acid (HCl e=1, H₂SO₄ e=2) and e_B (NaOH e=1, Mg(OH)₂ e = 2). Now carbonate and bicarbonate are a little tricky, but the mechanism for neutralization is that they both "neutralize" the HCl by forming H₂CO₃ (Of course this is an acidic end point, hence use of methyl red). HCO₃^- can neutralize one H⁺, so it has an e_B of 1 and carbonate has an e_B of 2 (e stands for equivalence. Note NH₃ has an e_B of 1)

All right, so how do you deal with the mixture? You don't! Just solve for V_A treating each bases separately and add the volumes of HCl needed to neutralize them. V_B is the same for each base.

Bicarbonate M_AV_A= M_BV_B and for Carbonate M_AV_A= 2M_BV_B

Good luck!

Answer 2

Look at the two balanced equations for the neutralization reactions:

NaHCO₃ + HCl ==> NaCl + CO₂ + H₂O

Na₂CO₃ + 2HCl ==> 2NaCl + CO₂ + H₂O

In the first case, 1 mol NaHCO₃ requires 1 mol HCl

In the 2nd case, 1 mol Na₂HCO₃ requires 2 mol HCl

So now we can find moles of HCl needed in each case as follows:

(1) 50.00 ml NaHCO₃ x 1 L/1000 ml x 0.02159 mol/L = 0.0010795 moles NaHCO₃ = 0.0010795 moles HCl

(2) 50.00 ml Na₂HCO₃ x 1 L/1000 ml x 0.0245 mol/L = 0.001225 mol Na₂HCO₃ = 0.00245 mol HCl (2x)

Total moles HCl needed = 0.0010795 mol + 0.00245 = 0.0035295 moles HCl needed

Now, since we know the molarity of the HCl is 0.06176 mol/L, we can calculate the volume needed:

0.06176 mol/L (x L) = 0.0035295 moles

x = 0.05715 L = 0.0572 L = 57.2 mls (to 3 significant figures)