J.R. S. answered 02/19/21
Ph.D. University Professor with 10+ years Tutoring Experience
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+