Question

Consider the insoluble compound copper(II) hydroxide , Cu(OH)2 . The copper(II) ion also forms a complex with ammonia . Write a balanced net ionic equation to show why the solubility of Cu(OH)2 (s) increases in the presence of ammonia and calculate the equilibrium constant for this reaction. For Cu(NH3)42+ , Kf = 6.8×1012 . Use the pull-down boxes to specify states such as (aq) or (s). + + K =

Answer 1

Cu(OH)2 is an insoluble complex. Ksp for Cu(OH)2 can be found using standard reference table.

Ksp for Cu(OH)2 is 2.2 x 10^-20

The ionization reaction can be written as

Cu(OH)2 (s) <-------------------> Cu ²⁺ (aq) + 2 OH^- (aq) Ksp = 2.2 x 10^-20 ...............Equation 1

The complexation reaction can be written as

Cu^2+ (aq) + 4 NH3 (aq) <-------------------> Cu(NH3)4 ^2+ (aq) Kf = 6.8 x 10^12 .......... Equation 2

Let's add equations 1 & 2

Cu(OH)2 (s) <-------------------> Cu ²⁺ (aq) + 2 OH^- (aq) Ksp = 2.2 x 10^-20

Cu^2+ (aq) + 4 NH3 (aq) <-------------------> Cu(NH3)4 ^2+ (aq) Kf = 6.8 x 10^12

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Cu(OH)2 (s) + 4 NH3 (aq) <-------------------> Cu(NH3)4^2+ (aq) + 2OH- Keq = Ksp * Kf

The net ionic equation is Cu(OH)2 (s) + 4 NH3 (aq) <-------------------> Cu(NH3)4^2+ (aq) + 2OH-(aq)

Keq = Ksp * Kf

Keq = 2.2 x 10^-20 * 6.8 x 10^12

Keq = 1.5 x 10^-7

The equilibrium constant for the net ionic equation is higher than the Ksp of Cu(OH)2 . This indicates that more product Cu(NH3)4^2+ is formed. Therefore we can say Cu(OH)2 is more soluble in NH3 and leads to formation of Cu(NH3)4^2+

Note : The value of equilibrium constant depends on Ksp value. Please check the value given to you and make the changes accordingly.