A solution is prepared by dissolving 23.8 g of ammonium sulfate in enough water

Question

A solution is prepared by dissolving 23.8 g of ammonium sulfate in enough water to make 100.0 mL of stock solution. A 11.50-mL sample of this stock solution is added to 51.10 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

NH4+     M SO42?     M

A solution is prepared by dissolving 23.8 g of ammonium sulfate in enough water to make 100.0 mL of stock solution. A 11.50-mL sample of this stock solution is added to 51.10 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.

Explanation / Answer

M of (NH4)2SO4 = 23.8g
V of soln = 100ml
V = 100/1000 = 0.1L
Mw of (NH4)2SO4 = 132.14g/mol
Nos of moles = 0.180moles
Conc = 0.180 / 0.1 = 1.8M
Then,
11.5ml aliquot of 0.53M was added to 51.1ml of water
Final vol of soln(2) = 11.5+51.1 = 62.6ml

C1V1 = C2V2
C2 = C1V1/V2
Where V1 = vol of aliquot taken from 100ml stock soln = 11.5ml; C1 = conc of stock = 1.8M; V2 = vol of soln(2) = 62.6ml
C2 = [1.8 x 11.5]/62.6
C2 = 0.3306

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