Question

a solution consists of 0.050 M Mg2+ and Cu2+. Which ion precipitates first when OH- is added? write out the equation for each and indicate the Ksp?

Answer 1

The Ksp values of Mg(OH)2 and Cu(OH)2 are known to be as ,

K_sp , Cu(OH)₂   = 4.8 x 10^-20

K_sp ,Mg(OH)2 = 5.61 x 10^-12

So , we need to write equations representing K_sp of the two salt solutions with given

ionic concentrations as 0.05 M for Cu²⁺ & Mg²⁺ respectively , and calculate the [ OH^- ] needed

to cause precipitation of hydroxide. According to solubility product principle , one requiring lesser

[OH^- ] such that its solubility product K_sp value is exceeded will get precipitated first.

Thus , For Cu(OH₂ )

...............Cu(OH)₂ (aq) <----------> Cu²⁺ (aq)   + 2OH^- (aq)

.................................K_sp = [ Cu²⁺ ] [OH^- ]²

................................[ OH^- ] = SQRT{ ( K_sp / [ Cu²⁺ ) }

..............................................= SQRT { 4.8 x 10^-20 / 0.05 )

..............................................= 9.7 979 x 10^-10 M

Similarly , for Mg(OH)₂   -

.............................................Mg (OH)₂ (aq) <---------------> Mg²⁺ (aq) + 2OH^-

.......................................& K_sp  = [ Mg ²⁺ ] [ OH^- ]²

.........................................[ OH^- ] = SQRT { ( K_sp / [ Mg²⁺ ) }

.................................................... = SQRT { (5.61 x 10^-12 / 0.05 ) }

......................................................= 1.0593 x 10^-7 M

Now, since a lower value of [ OH^- ] is required to exceed the K_sp value of Cu(OH)₂ it will get

precipitated first , as per solubility product principle.