Question

Using the appropriate Ksp values, find the concentration of Cu2+ ions in the solution at equilibrium after 600 mL of 0.50 M aqueous Cu(NO3)2 solution has been mixed with 300 mL of 0.20 M aqueous KOH solution. (Enter in M.) (Ksp for Cu(OH)2 is 2.6x10-19).

Now find the concentration of OH− ions in this solution at equilibrium. (Enter in M.)

Answer 1

no of moles of Cu(NO3)2   = molarity * volume in L

                                            = 0.5*0.6 = 0.3 moles

Cu(NO3)2 ------------------> Cu⁺² + 2NO3^-

0.3M                                   0.3M

no of moles KOH           = molarity * volume in L

                                      = 0.2*0.3   = 0.06 moles

KOH is limiting reagent

KOH ---------------------> K⁺ + OH^-

0.06moles                              0.06 moles

Cu⁺² + 2OH^- --------------> Cu(OH)2

2 moles of OH^- react with 1 mole of Cu⁺²

0.06moles of OH^- react with = 0.06*1/2 = 0.03 moles of Cu⁺²

no of moles of remaining Cu⁺² = 0.3-0.03   = 0.27moles

molarity of Cu⁺²                       = no of moles/volume in L

                                                  = 0.27/0.9   = 0.3M

Cu(OH)2 ----------> Cu⁺² + 2OH^-

Ksp    = [Cu⁺²][OH^-]²

2.6*10^-19   = 0.3*[OH^-]²

[OH-]²         = 2.6*10^-19/0.3   = 8.6*10^-19

[OH-]            = 9.3*10^-9 M