Question

If an impure sample of soda ash is known to contain 40.00% of Na2CO3, calculate the sample size in mg which will react with exactly 40.00 mL of 0.1100 M HCl.What is the % purity of a soda ash sample if it took 42.15mL of 0.1100 M HCl to titrate 0.5017 g of the sample?

Answer 1

The sample size is 233.19 mg and the purity of the sample is 48.97%

Why?

The chemical reaction between Na₂CO₃ and HCl is

Na₂CO₃ + 2HCl → 2NaCl + H₂O + CO₂

The sample size that will react with 40 mL of 0.1100 M HCl is found by applying the following conversion factor to go from L of solution of HCl to mg of Na₂CO₃:

$0.04 Lsolution*\frac{0.1100molesHCl}{1L solution}*\frac{1 mole Na_2CO_3}{2 moles HCl}*\frac{105.99 gNa_2CO_3}{1 mole Na_2CO_3}*\frac{1000mg}{1 g}\\ \\=233.19mgNa_2CO_3$

Now, in order to find the percentage purity we have to apply another conversion factor to go from mL of solution of HCl to g of Na2CO3, and dividing everything over the mass of the soda ash sample, followed by multiplying the result times 100 to find the percentage:

$0.04215 Lsolution*\frac{0.1100molesHCl}{1L solution}*\frac{1 mole Na_2CO_3}{2 moles HCl}*\frac{105.99 gNa_2CO_3}{1 mole Na_2CO_3}\\ \\=0.2457mgNa_2CO_3*\frac{100\%}{0.5017 g} = 48.97 \%$

Have a nice day!

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