Question

20.0 mL of 0.100 M copper(II) sulfate reacts with 20.0 mL of 0.100 M sodium hydroxide to form a precipitate (Cu(OH) 2 2 ). What is the concentration (molarity) of Cu 2+ in the resulting solution?

Answer 1

moles Cu2+ = 0.1 M x 20 ml

                    = 2 mmol

moles OH- = 0.1 M x 20 ml

                  = 2 mmol

moles Cu(OH)2 formed = 1 mmol

molarity of Cu2+ in resulting solution = 1 mmol/40 ml

                                                           = 0.025 M