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)

Part A

What volume of 0.190 M HCl is required for the complete neutralization of 1.70 g of NaHCO3 (sodium bicarbonate)?

Express your answer to three significant figures and include the appropriate units.

Part B

What volume of 0.120 M HCl is required for the complete neutralization of 1.80 g of Na2CO3 (sodium carbonate)?

Express your answer to three significant figures and include the appropriate units.

Part C

A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.170 g 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?

Express your answer to three significant figures and include the appropriate units.

Answer 1

Balanced equation:
NaHCO₃(aq) + HCl(aq) = H₂CO₃(aq) + NaCl(aq)
Reaction type: double replacement

Part A

1.70 g of NaHCO3 (Molar mass is 84) = 1.7 / 84 = 0.020236 Moles

0.020236 Moles of NaHCO3 need the same moles of HCl

VOlume of 0.190 HCl = 0.020236 x 1000 / 0.19 = 106.5 ml

Hence 106.505 ml of 0.19M HCl is need.

Part B

Balanced equation:
Na₂CO₃(aq) + 2 HCl(aq) = H₂CO₃(aq) + 2 NaCl(aq)
Reaction type: double replacement

1.80 g of Na2CO3 (Molar mass is 105.988 ) =   1.8 / 105.988 = 0.016982 Moles

According to the above equation 0.016982 Moles Na2CO3 needs ( 0.016982 x 2) 0.033964 Moles of HCl

0.033964 Moles of HCl = 0.033964 x 1000 / 0.12 = 283.03 ml

283.03 ml of HCl is need to neutralization of 1.80 g of Na2CO3

Part C

23.98 mL of 0.100 M HCl = 23.98 x 0.1 x 40 /1000 = 0.09592 gm NaOH presents

0.700 mL of 0.100 M HCl   = 0.7 x 0.1 x 2 x 105.988 /1000 = 0.014838 gm Na2CO3 presents

Mass percentage of Na2CO3 = 0.014838 X 100 / 0.09592 = 15.45 %