chem ~molarity queation

Question

2.65g of anhydrous sodium carbonate was dissolved in water made up to 250.0 cm^3 solution 25.0 cm^3 of this solution required 26.3 cm^3 of a hydrochloric acid soltion for complete neutralization . Find the molarity of the hydrochloric acid.

Answer 1

Molar mass of Na2CO3 = 23x2 + 12 + 16x4 = 106 g mol-1
No. of moles of Na2CO3 = mass/(molar mass) = 2.65/106 = 0.025 mol
Molarity of Na2CO3(aq) = mol/V = 0.025/(250/1000) = 0.1 mol dm-3

Consider the titration :
Na2CO3 + 2HCl → 2NaCl + H2O + CO2
Mole ratio Na2CO3 : HCl = 1 : 2
No. of moles of Na2CO3 used = MV = 0.1 x (25/1000) = 0.0025 mol
No. of moles of HCl used = mol/V = 0.0025 x 2 = 0.005 mol
Molarity of HCl = mol/V = 0.005/(26.3/1000) = 0.190 mol dm-3

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They key point of the calculation is that NOT all Na2CO3 solution. Out of 250 cm3 (i.e. 0.025 mol) of the total solution made, only 25 cm3 (i.e. 0.0025 mol) of Na2CO3 is used in the titration.

Answer 2

molar mass: Na = 23g, C = 12.0g, O = 16g

Molar mass of sodium carbonate = 23x2+ 12+16x3 = 106g
no. of mol of sodium carbonate = 2.65 / 106 = 0.025 mol

Na2CO3 + 2 HCl ----> 2 NaCl + H2O + CO2

(no. of mol of sodium carbonate )/ (no. of mol of HCl) = 0.5

no. of mol of HCl = 2 (no. of mol of sodium carbonate) = 0.025 x 2 = 0.05 mol

Molarrty of HCl = no. of mol of sodium carbonate / vol of carbonate used in dm3
= 0.05 / (26.3 / 1000)
= 1.90M