What volume of 0.011 M HCl is required to titrate a solution containing 257 mL of 0.011 M Na2CO3 and 257 mL of 0.011 M HCO3- ?

Question

The common-ion effect plays a key role in controlling the pH of solutions that contain both a weak acid and its conjugate base.

Answer 1

Answering this question is a little tricky because there is no information about which indicator was used for this titration.

For example:

If the Phenophthalene was used as an indicator, it changed the color when the following reaction step was completed.

Na₂CO₃ + HCl -----> NaHCO₃ + NaCl

If Methyl Orange was used as an indicator then it changed color at the endpoint of all HCO3 - completed the reaction with HCl shown below.

NaHCO₃ + HCL --------> NaCl + H₂O + CO₂

I presume here methyl orange (MO) was used as an indicator; Hence it will change color when Na₂CO₃ and HCO₃^- completely react with HCl!

So if we write the reaction equation between Na₂CO₃ and HCL in the presence of MO as an indicator;

Na₂CO₃ + 2HCl -----> 2NaCl + H₂O + CO₂

Mole of Na₂CO₃ in the given solution (n) = cv

= 0.011 mol/L * 257 * 10-³ L

= 2.827 * 10^-3 mol

Required HCl mole to completely react with Na₂CO₃ = 2.827 * 10^-3 mol * 2 (nNa₂CO₃ : nHCl =1:2)

= 5.654 * 10^-3 mol

HCO₃^- + HCl ----> Cl^- + CO₂ +H₂O

Mole of HCO₃^- in the given solution (n) = cv

= 0.011 mol/L * 257 * 10-³ L

= 2.827 * 10^-3 mol

Required HCl mole to completely react with HCO₃^- = 2.827 * 10^-3 mol *1 (nHCO₃^- : nHCl =1:1)

Required total HCl mol = (5.654 * 10^{-3 +} 2.827 * 10^-3) mol

= 8.481 * 10^-3 mol

Required 0.011 M HCl = 8.481 * 10^-3 mol/ 0.011 molL^-1

= 771 * 10 -3 L or 771 mL

Answer 2

Titration of Na₂CO₃ with HCl:

Na₂CO₃ + 2HCl ==> 2NaCl + H₂O + CO₂ ... balanced equation

moles Na₂CO₃ present = 257 ml x 1 L / 1000 ml x 0.011mol / L = 2.827x10^-3 mols

moles HCl needed = 2.827x10^-3 mols Na₂CO₃ x 2 mol HCl / mol Na₂CO₃ = 5.65x10^-3 mols HCl

Titration of HCO₃^- with HCl:

HCO₃^- + HCl ==> Cl^- + H2O + CO₂ ... balanced equation

moles HCO₃^- = 257 ml x 1 L / 1000 ml x 0.011 mol / L = 2.827x10^-3 mols

moles HCl needed = 2.827x10^-3 mols HCO₃^- x 1 mol HCl / mol HCO₃^- = 2.827x10^-3 mols HCl

Total moles HCl needed = 5.65x10^-3 + 2.83x10^-3 = 8.48x10^-3 moles HCl needed

Volume of HCl = 8.48x10^-3 mols / L x 1 L / 0.011 mol = 0.771 L (771 mls)