Question

1. Calculate the molarity of an HCl solution if a volume of 39.54 mL of this solution is required to titrate 0.2348 g Na2CO3. (Bromocresol green is used as an indicator).
2. This HCl solution (calculated in question 1.) is then used to analyze an unknown sample. Calculate the % Na2CO3 in the sample if a volume of 23.44 mL of the HCl solution is required to titrate 0.4089 g of the sample

Answer 1

Answer =1

Given volume of HCl solution = 39.54 mL ( i.e. in L = 39.54 / 1000 = 0.03954 L)

Mass of Na₂CO₃ = 0.2348 g & molar mass of Na₂CO₃ = 105.98 g/mol

Now calculate mole of Na₂CO₃ = 0.2348 g / 105.98 g/mol =0.002216 moles

Number mole of Na₂CO₃ = 0.002216 moles

We know the reaction

1 Na2CO3 (s) + 2 HCl (aq) → 2 NaCl (aq) + 1 CO2 (g) + 1 H2O (l)

1 mole of Na₂CO₃ react with 2 mole of HCl so number of moles of HCl will be

= 2 x 0.002216

=0.004432 moles of HCl

Now molarity of HCl = number of moles of HCl / volume in L = 0.004432 moles / 0.03954 L

Molarity of HCl = 0.1121 M

Answer = 2

Given

Molarity of HCl = 0.1121 M, Volume of HCl solution = 23.44 mL ( i.e. in L = 23.44 / 1000 = 0.02344 L) , mass of unknown sample = 0.4089 g

Now calculate Number of moles of HCl = molarity x volume in Liter = 0.1121 mol/L x 0.02344 L

Number of moles of HCl = molarity x volume in Liter = 0.1121 mol/L x 0.02344 L =0.002628 moles

Number of moles of HCl = 0.002628 moles

Now 1 mole of Na₂CO₃ reacts with 2 mole of HCl so moles of Na₂CO₃ will be

Number moles of Na₂CO₃ = Number of moles of HCl / 2 = 0.002628 / 2

Number moles of Na₂CO₃ = 0.001314 moles

Now mass of Na₂CO₃ = 0.001314 mole x 105.98 g/mol

Now mass of Na₂CO₃ = 0.001314 mole x 105.98 g/mol

mass of Na₂CO₃ = 0.1392 g

now calculate % mass of Na₂CO₃ if we have 0.1392 g mass of Na₂CO₃ in 0.4089 g unknown sample

% mass of Na₂CO₃ = (actual mass of Na₂CO₃ / mass of unknown sample) x 100

% mass of Na₂CO₃ =30.04 %