Question

When 18.0 mL of a 5.50×10^-5M sodium hydroxide solution is combined with 15.0 mL of a 6.54×10^-4M cobalt(II) iodide solution does a precipitate form?  (yes or no)

For these conditions the Reaction Quotient, Q, is equal to?

Answer 1

The balanced reaction formed when sodium hydroxide solution is combined with cobalt(II) iodide solution is

2NaOH(aq) + CoI₂(aq) ----> Co(OH)₂(s) + 2NaI (aq)

K_sp for Co(OH)₂ is 5.5×10^–6

Number of mmoles of NaOH , n = Molarity x volume in mL

                                           = 5.50×10^-5M x 18.0 mL

                                           = 0.0099 moles

Number of moles of CoI2 , n' = Molarity x volume in mL

                                         = 6.54×10^-4M x 15.0 mL

                                        = 0.0098 moles

From the balanced reaction,

1 mole of CoI2 reacts with 2 moles of NaOH

M moles of CoI2 reacts with 0.0099 moles of NaOH

M = ( 1x0.0099) / 2

   = 0.00495 moles of CoI2

Therefore 0.0098 - 0.00495 = 0.00485 moles of CoI2 left unreacted

So NaOH is the limiting reactant.

Again 2 moles of NaOH produces 1 moles of Co(OH)2

      0.0099 moles of NaOH produces 0.0099/2= 0.00495 moles of Co(OH)2

Concentration of Co(OH)2 = number of moles / volume of the mixture

                                     = 0.00495 mol / ( (18.0+15.0)mL x ( 10^-3 L/mL))

                                    = 0.0165 M

Co(OH)2 ---> Co2+ + OH-

[Co²⁺] = 0.0165 M

[OH^-] = 2x0.0165 M = 0.033 M

Reaction quotient for Co(OH)2 is , Q = [Co²⁺] [OH^-]²

                                                     = 1.79*10^-5

Since Q > K_sp precipitate will be formed