What is the concentration of Cu^{2+} in the resulting solution?

Question

23.6 mL of 0.141 M copper(II) sulfate reacts with 15.9 mL of 0.114 M sodium hydroxide to form a precipitate (Cu(OH)). What is the concentration of Cu in the resulting solution?

Thank you!

Answer 1

CuSO₄(aq) + 2NaOH(aq) ==> Cu(OH)₂(s) + Na₂SO₄(aq) ... balanced equation

moles CuSO₄ = 23.6 ml x 1 L/1000 ml x 0.141 mol/L = 0.003328 moles

moles NaOH = 15.9 ml x 1 L/1000 ml x 0.114 mol/L = 0.001813 moles - LIMITING REACTANT

Moles Cu(OH)₂ precipitated = 0..001813 mol NaOH x 1 mol Cu(OH)₂ / 2 mol NaOH = 9.1x10^-4 moles

Moles Cu(OH)₂ remaining is solution = 3.3x10^-3 - 9.1x10^-4 = 0.00239 moles CuSO₄ remaining

Moles Cu²⁺ in solution = 0.00239 moles

Total volume = 23.6 ml + 15.9 ml = 39.5 ml = 0.0395 L

Final concentration of Cu²⁺ in solution = 0.00239 moles/0.0395 L = 0.0611 M Cu²⁺