methyl red end point detection

Question

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

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)