Question

Using the appropriate K_sp values, find the concentration of Cu²⁺ ions in the solution at equilibrium after 650 mL of 0.45 M aqueous Cu(NO₃)₂ solution has been mixed with 500 mL of 0.25 M aqueous KOH solution. (Enter in M.) (K_sp for Cu(OH)₂ is 2.6x10^-19).AND find the concentration of OH^? ions in this solution at equilibrium. (Enter in M.)

Answer 1

Concentration of Cu(NO₃)₂ solution = 0.45 M = 0.45 mol L^-1

Volume of Cu(NO₃)₂ solution used = 650 mL = 0.650 L -------[1 ml =0.001 L]

No. of moles of Cu(NO₃)₂  in 650 mL Cu(NO₃)₂ solution = ( 0.45 mol L^-1) * (0.650 L) =0.2925 moles

1 mole of Cu(NO₃)₂ contains 1 mole of Cu²⁺ ions.

Therefore, no.of moles of Cu²⁺ ions added = 0.2925 moles

Concentration of KOH solution = 0.25 M = 0.25 mol L^-1

Volume of KOH solution used = 500 mL = 0.500 L -------[1 ml =0.001 L]

No. of moles of KOH in 500 mL KOH  solution = ( 0.25 mol L^-1)*(0.500 L ) = 0.125 moles

1 mole of KOH contains 1 mole of OH ^- ions.

Therefore, no.of moles of OH ^-  ions added = 0.125 moles

The equation for formation of Cu(OH)₂ is

Cu²⁺ (aq) + 2 OH^-(aq) Cu (OH)₂ (s)

Therefore, 1 mole of Cu²⁺ ions requires 2 moles of OH^- ions to react completely.

That is, 2 moles of OH^- ions react with 1 mole of  Cu²⁺ ions.

Therefore, No. of moles of Cu²⁺ ions that will react with 0.125 moles of OH^- ions will be = (1/2 ) * 0.125 moles

= 0.0625 moles

Therefore, 0.0625 moles of Cu²⁺ ions will react with available  OH^- ions, and the remaining will be in the solution.

No. of moles of Cu²⁺ ions remaining in the solution

=No. of moles of  Cu²⁺ ions added - No. of moles of  Cu²⁺ ions reacted with OH^- ions  

=0.2925 moles - 0.0625 moles

= 0.23 moles

Final volume of solution = volume of  Cu(NO₃)₂ solution + volume of KOH solution

= 650 mL + 500 mL

=1150 mL

= 1.150 L

Concentration of Cu²⁺ ions remaining in the solution

= No. of moles of Cu²⁺ ions remaining in the solution / Final volume of solution(in L)

=0.23 moles / 1.150 L

=0.2 M

Now, K_sp of Cu (OH)₂ (s) will be given by

K_sp = [Cu²⁺][OH^-]²

Given,

K_sp for Cu(OH)₂ = 2.6x10^-19

Therefore,

K_sp = [Cu²⁺][OH^-]²

2.6x10^-19 = ( 0.2 )[OH^-]²

[OH^-]² = 1.3*10^-18

[OH^-] = 1.14 *10^-9 M

Therefore,

Concentration of Cu²⁺ ions remaining in the solution  = 0.2 M

Concentration of OH^- ions in this solution at equilibrium = 1.14 *10^-9 M