Question

Part A: What volume of 0.100 M HCl is required for the complete neutralization of 1.50 g of NaHCO3 (sodium bicarbonate)?

Part B: What volume of 0.140 M HCl is required for the complete neutralization of 1.30 g of Na2CO3 (sodium carbonate)?

Part C: A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.200 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?

I don't understand how to do this question at all. Please and thank you show and explain how you got the answers!

Answer 1

A. HCl + NaHCO3 NaCl + CO2 + H2O

Thus, 1 mole NaHCO3 requires 1 mol HCl to undergo complete neutralisation.

Now,

Moles of NaHCO3 = Given Wt / Mol Wt

= 1.5 g / 87 g/mol

= 0.017 mol

Thus,

Mol of HCl = 0.017 mol

Now,

Mol of HCl = Molarity * Volume

0.017 mol = 0.1 M* V

V = 0.17 L= 170 mL

Thus,

170 mL HCl is needed

B. 2 HCl + Na2CO3 2 NaCl + CO2 + H2O

Thus, 1 mole Na2CO3 requires 2 mol HCl to undergo complete neutralisation.

Now,

Moles of Na2CO3 = Given Wt / Mol Wt

= 1.3 g / 106 g/mol

= 0.012 mol

Thus,

Mol of HCl = 0.012*2 mol = 0.024 mol

Now,

Mol of HCl = Molarity * Volume

0.024 mol = 0.14 M* V

V = 0.17 L= 170 mL

Thus,

170 mL HCl is needed.