Question

What is the cobalt (II) ion concentration in a solution prepared by mixing 395 mL of 0.381 M cobalt (II) nitrate with 463 mL of 0.204 M sodium hydroxide? The Ksp of cobalt (II) hydroxide is 1.3 × 10-15

*** Someone said the answer was 1.2x10^{-6 ,}1.2x10^-14, , 6.9x10^-14 and its incorrect. HELPP!!

Answer 1

Number of moles of cobalt (II) nitrate , n= molarity x volume in L

                                                         = 0.381 M x 0.395 L

                                                         = 0.150 mol

Number of moles of sodium hydroxide , n' = molarity x volume in L

                                                             = 0.204 M x 0.463 L

                                                            = 0.094 mol

The balanced reaction is : Co(NO₃)₂ + 2NaOH Co(OH)₂ + 2NaNO₃

According to the balanced equation ,

1 mole of Co(NO₃)₂ reacts with 2 moles of NaOH

M mole of Co(NO₃)₂ reacts with 0.094 moles of NaOH

M = (1x0.094) / 2

   = 0.047 moles

So 0.150 - 0.047 = 0.103 moles of Co(NO₃)₂ left unreacted

The concentration of Co(NO₃)₂ left unreacted is C' = number of moles left / total volume in L

                                                                        = 0.103 mol / (0.395+0.463)L

                                                                        = 0.120 M

From the balanced equation ,

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

0.094 moles of NaOH produces N mole of Co(OH)₂   

N = ( 1x0.094)/ 2

   = 0.047 moles

So concentration of Co(OH)₂ is C= number of moles / total volume in L

                                                 = 0.047 mol / (0.395+0.463)L

                                                 = 0.055 M

Therefore the total concentration of Co(II) = Concentration of Co(II) in Co(NO₃)₂ +Concentration of Co(II) in Co(OH)₂

                                                            = 0.120 + 0.055 M

                                                            = 0.175 M