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. At neutralization point, no. of moles of NaHCO3 will be equal to no. of moles of HCl

no. of moles of NaHCO3 = given mass/molar mass = 1.5g/ 84g/mol = 0.0179moles

no. of moles of HCl = molarity x volume in liters = 0.18 x V1

So, At neutralization, 0.0179 moles = 0.18 x V1

V1 = 0.0992L = 99.2mL

2.

no. of moles of NaHCO3 = given mass/molar mass = 1.1g/ 84g/mol = 0.013moles

no. of moles of HCl = molarity x volume in liters = 0.12 x V1

So, At neutralization, 0.013 moles = 0.12 x V1

V1 = 0.1091L = 109.13mL