Question

Using the appropriate Ksp values, find the concentration of NO3− ions in the solution at equilibrium after 700 mL of 0.35 M aqueous Cu(NO3)2 solution has been mixed with 350 mL of 0.40 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

2KOH + Cu(NO3)2 ---> Cu(OH)2 + 2KNO3

Concentration of NO3- ions in equilibrium:

The only thing that NO3- does is moving from Cu(NO3)2 to form KNO3, it doesn`t precipitate, so the way to find its concentration is:

mol Cu(NO3)2= 0.7L x 0.35M= 0.245 mol

1 mol of Cu(NO3)2 contains 2 moles of NO3-

mol NO3-= 2 x 0.245mol= 0.49 mol

Final volume after mixing= 700 + 350= 1050mL= 1.05L

[NO3-]= 0.49mol/1.05L= 0.467M

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Now [OH-]:

mol of KOH= 0.35L x 0.40M= 0.14 mol

mol of Cu(NO3)2= 0.245 mol

2 mol KOH ----------- 1 mol Cu(NO3)2

0.14 mol KOH --------- x= 0.07 mol Cu(NO3)2---> we have 0.245mol, so it is in excess, KOH is the limiting reactant.

So, we have that 0.07 mol of Cu+2 will precipitate, the rest will reamin in solution.

Cu+2 free= 0.245 mol - 0.07 mol= 0.175 mol

[Cu+2]free= 0.175mol/1.05L= 0.167M

Cu(OH)2 <-----> Cu+2 + 2OH-

Ksp= [Cu+2][OH-]²= 0.167 x[OH-]²

[OH-]= 1.25x10^-9M