contestada

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

Respuesta :

Answer:

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

Otras preguntas