The concentration of Cu2 + ions in the water (which also contains sulfate ions) discharged from a certain industrial plant is determined by adding excess sodium sulfide (Na2S) solution to 0.800 L of the water. The molecular equation is Na2S (aq) + CuSO4 (aq) → Na2SO4 (aq) + CuS (s) Calculate the molar concentration of Cu2 + in the water sample if 0.0183 g of solid CuS is formed

0.000239 M

Explanation:

Na2S (aq) + CuSO4 (aq) → Na2SO4 (aq) + CuS (s)

molar mass of = 159.609 g/mol

molar mass of CuS = 95.611 g/mol

if 159.609 g/mol yielded 95.611 g/mol of CuS

then x g of CuSO4 that yielded 0.0183 g = 159.609 * 0.0183 / 95.611 = 0.0305 g

molar concentration = mole / volume in liters

number of mole of CuSO4 in the water sample = 0.0305 g / 159.609 g/mol = 0.000191 moles

molar concentration of CuSO4 = 0.000191 moles / 0.8 L = 0.000239 M

since when 1 mole of CuSO4 will dissociate to produce 1 mole of Cu2+

the molar concentration Cu2 + = 0.000239 M