Question

A solution is prepared by dissolving 12.7 g ammonium sulfate in enough water to make 124.0 mL of stock solution. A 13.70-mL sample of this stock solution is added to 56.50 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Answer 1

Answer: a) Ammonia — 0.3025M

b) sulphate — 0.1512M

Step-by-step explanation:

Molarity = mol solute / L solution

Mass of the solution = 12.7 x (1/132.14) = 0.0961 mol in 124mL

Molarity = 0.0961/0.124 L solution = 0.775M

So 13.70 ml of this solution is added to 56.50 mL we have 70.2 mL total solution

Now using dilution equation

M1V1 = M2V2

To find the molarity of the new 70.2mL solution

(0.775M x 13.7 ) = M2 x 70.2

M2 = (0.775M x 13.7 ) / 70.2 = 0.1512M

Therefore there’s 0.1512 moles ammonia sulphate per liter of the solution

Therefore the concentration of Ion is

1. 2 x 0.1512 = 0.3025M of ammonia

2. 1 x 0.1512 = 0.1512M of sulphate