Answer: molarity of ammonium ions = 0.274mol/L
molarity of sulfate ions = 0.137mol/L
<em>Note: The complete question is given below</em>
A solution is prepared by dissolving 10.8 g ammonium sulfate in enough water to make 100.0 mL of stock solution. A 10.00-mL sample of this stock solution is added to 50.00 mL of water. Calculate the concentration of ammonium ions and sulfate ions in the final solution.
Explanation:
Molar concentration = no of moles/volume in liters
no of moles = mass/molar mass
mass of ammonium sulfate = 10.8g, molar mass of ammonium sulfate, (NH₄)₂SO₄ = (14+4)*2 + 32+ (16)*4 = 132g/mol
no of moles = 10.8g/132g/mol = 0.0820moles
<em>Molarity of stock solution = 0.0820mol/(100ml/1000ml* 1L) = 0.0820mol/0.1L Molarity of stock solution = 0.820mol/L</em>
Concentration of final solution is obtained from the dilution formula,
<em>C1V1 = C2V2</em>
C1 = 0.820M, V1 = 10mL, C2 = ?, V2 = 60mL
C2 = C1V1/V2
C2 = 0.820*10/60 = 0.137mol/L
molar concentration of ions = molarity of solution * no of ions
molarity of ammonium ions = 0.137mol/L * 2 = 0.274mol/L
molarity of sulfate ions = 0.137 mol/L * 1 = 0.137mol/L
