Question

Carbonic acid (H2CO3) is a weak diprotic acid with Ka1=4.43�10?7 and Ka2=4.73�10?11. When sodium bicarbonate (NaHCO3) is titrated with hydrochloric acid (HCl), it acts as a weak base according to the equation

NaHCO3(aq)+HCl(aq)?H2CO3(aq)+NaCl(aq)

1)What volume of 0.180M HCl is required for the complete neutralization of 1.50g of NaHCO3 (sodium bicarbonate)?

2)What volume of 0.120M HCl is required for the complete neutralization of 1.10g of Na2CO3 (sodium carbonate)?

3)A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.160g of this sample requires 23.98 mL of 0.100 M HCl. An additional 0.700 mL of 0.100 M HCl is required to reach the methyl orange endpoint. What is the percentage of Na2CO3 by mass in the sample?

Answer 1

1.Given the weight of NaHCO3 = 1.50g

Moles of NaHCO3 = weight of NaHCO3 / molecular mass = 1.50g / 84gmol^-1 = 0.0179 mol

Given concentration of HCl = 0.180M

Let the volume of HCl be ‘V’ L

Hence moles of HCl = MV = 0.180V mol

The given chemical reaction is

NaHCO3(aq)+HCl(aq) -- > H2CO3(aq)+NaCl(aq)

1 mol                1 mol              1 mol            1 mol

In the above chemical reaction,

Stoichiometric ratio for NaHCO3 to HCl = 1/1

Hence Reacting moles of HCl = reacting moles of NaHCO3

= > 0.180V mol = 0.0179 mol

= > V = 0.0179 / 0.180 = 0.0994 L = 99.4 mL (answer)

1. Given the weight of Na2CO3 = 1.10g

Moles of Na2CO3 = weight of Na2CO3 / molecular mass = 1.10g / 106gmol^-1 = 0.0104 mol

Given concentration of HCl = 0.120M

Let the volume of HCl be ‘V’ L

Hence moles of HCl = MV = 0.120V mol

When Na2CO3 reacts with HCl, the following reaction occurs.

Na2CO3(aq)+2HCl(aq) -- > H2CO3(aq)+2NaCl(aq)

1 mol                2 mol              1 mol            2 mol

In the above chemical reaction,

Stoichiometric ratio for Na2CO3 to HCl = 1/2

reacting moles of Na2CO3 = 2x(Hence Reacting moles of HCl )

= > 0.0104 mol = 2x(0.120V) mol = 0.240V mol

= > V = 0.0104 / 0.240 = 0.0433 L = 43.3 mL (answer)