Question

Numerical: How will you prepare a 0.5 N `Na_2CO_3` solution having a volume of 150 ml?

Answer 1

You will add enough water to 3.98 g of `"Na"_2"CO"_3` to make a total of 150 mL.

Explanation:

Step 1. Write the chemical equation for the reaction with HCl.

`"Na"_2"CO"_3 + "2HCl" → "2NaCl" + "H"_2"O" + "CO"_2`

Step 2. Calculate the mass of 1 equivalent of `"Na"_2"CO"_3`.

`"1 eq Na"_2"CO"_3 = 1 cancel("mol HCl") × (1 cancel("mol Na₂CO₃"))/(2 cancel("mol HCl")) × ("105.99 g Na"_2"CO"_3)/(1 cancel("mol Na₂CO₃")) = "53.00 g Na"_2"CO"_3`

Step 3. Calculate the number of equivalents required.

`"Eq of Na"_2"CO"_3 = 0.150 cancel("L soln") × ("0.5 eq Na"_2"CO"_3)/(1 cancel("L soln")) = "0.075 eq Na"_2"CO"_3`

Step 4. Calculate the mass of `"Na"_2"CO"_3` required.

`"Mass of Na"_2"CO"_3 = 0.075 cancel("eq Na₂CO₃") × ("53.00 g Na"_2"CO"_3)/(1 cancel("eq Na₂CO₃")) = "3.98 g Na"_2"CO"_3`