Question

1. What volume of 0.200 M HCl is required for the complete neutralization of 1.20 g of Na2CO3 (sodium carbonate)?

2. A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.130 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 HAVE ONE ATTEMPT REMAINING PLEASE DO WELL)

Answer 1

1. What volume of 0.200 M HCl is required for the complete neutralization of 1.20 g of Na2CO3 (sodium carbonate)?

The reaction: Na2CO3 + HCl = H2O + CO2 + NaCl

balance: Na2CO3 + 2HCl = H2O + CO2 + 2NaCl

mol of Na2CO3 = mass/MW = 1.2/105.9888 = 0.01132 mol of Na2CO3

now..

mol of HCl required = 2x0.01132 = 0.02264 mol of HCl

M = mol/V

V mol/M

V = 0.02264 / 0.20

V = 0.1132 L

V= 113.2 mL