What volume of 0.140 M HCl is required for the complete neutralization of 1.40g

Question

What volume of 0.140M HCl is required for the complete neutralization of 1.40g of NaHCO3 (sodium bicarbonate)?

What volume of 0.200M HCl is required for the complete neutralization of 1.30g of Na2CO3 (sodium carbonate)?

A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.150g 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?

Explanation / Answer

Q1.

What volume of 0.140M HCl is required for the complete neutralization of 1.40g of NaHCO3 (sodium bicarbonate)?

the reaction:

NaHCO3 + HCl --> H2O + CO2 + NaCl

so..

mol of NaHCO3 = mass/MW = 1.4/84 = 0.016666 mol of NaHCO3

so 0.016666 mol of HCl

M = mol/V

V = mol/M

V = 0.016666/0.140 = 0.119 L = 119 mL

What volume of 0.200M HCl is required for the complete neutralization of 1.30g of Na2CO3 (sodium carbonate)?

pretty similar:

mol of NaHCO3 = mass/MW = 1.4/84 = 0.016666 mol of NaHCO3

so 0.016666 mol of HCl

M = mol/V

V = mol/M

V = 0.016666/0.20= 0.08333 L = 83.33 mL

A sample of NaOH (sodium hydroxide) contains a small amount of Na2CO3 (sodium carbonate). For titration to the phenolphthalein endpoint, 0.150g 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?

Na2CO3 + NAOH...

initial H+ used = MV = 23.98*0.1 = 2.398 mmol of H+

this was used for NaOH and Na2CO3

now, mmol of H+ extra = MV = 0.7*0.1 = 0.07 mmol of H+

so

mmol of Na2CO3 --> 0.07 mmol

mass = mmol*M W= 0.07*105.9888 = 7.41921 mg of Na2CO3

mass of NaOH --> 150 mg - 7.41921 mg = 142.58 mg of NaOH

% Na2CO3 = mass of Na2CO3 /total mass * !00% = 7.41921 /150 *100 = 4.9461 %

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