A 1.9450 g sample containing only Na2CO3 and NaHCO3 is titrated with 35.31 mL of 0.8724...

A 1.9450 g sample containing only Na₂CO₃ and NaHCO₃ is titrated with 35.31 mL of 0.8724 M HCl. Calculate the percentage sodium carbonate and sodium bicarbonate.

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Expert Solution

Let mass of Na2CO3 be x gm and mass of NaHCO3 = 1.945-x gm
Molar mass of Na2CO3 = 106 g/mol
Molar mass of NaHCO3 = 84 g

number of moles of Na2CO3 = mass /molar mass = x/106 mol
number of moles of NaHCO3= mass /molar mass = (1.945-x)/84 mol

equivalence of Na2CO3 = 2* x/106 {since it has 2 Na+}
equivalence of NaHCO3= 1*(1.945-x)/84

Number of equivalence of HCl = number of moles of HCl
= M*V
= 0.8724 M * (0.03531 L)
=0.0308

equivalent of Na+ = equaivalent of HCl
2* x/106 + (1.945-x)/84 = 0.0308
0.01887x + 0.0232 - 0.0119x = 0.0308
0.00697 x = 0.0076
x= 1.09 g
mass of sodium carbonate = 1.09 g
mass of sodium bicarbonate = 1.945 - 1.09 =0.855 g

percentage sodium carbonate = mass of sodium carbonate *100/ total mass
= 1.09*100 / 1.945
= 56 %
percenatge of sodium bicarbonate = 100 - 56% = 44%

queen_honey_blossom answered 3 weeks ago

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