A solution is prepared by mixing 50.0 ml of 0.50 m cu (no3) 2 with 50.0 ml of 0.50 m co (no3) 2. sodium hydroxide is slowly added to the mixture. which precipitates first? [ksp (cu (oh) 2) = 2.2 * 10-20, ksp (co (oh) 2) = 1.3 * 10-15]

When we dilute both molarities of Cu (NO3) and CO (NO3) 2 to half

to become 0.25 M Cu (NO3) & 0.25 M CO (NO3) 2

and when the value of Ksp CO (OH) 2 = 1.6 x 10^-15

and the value of Ksp Cu (OH) 2 = 2.2 x 10^-20

1) so, for CO (OH) 2 Ksp expression = [CO+2][OH-]^2

by substitution:

1.6 x 10^-15 = 0.25 m * [OH-]^2

∴ [OH] = 8 x 10^-8

∴POH = - ㏒[OH-]

= - ㏒ 8 x 10^-8

= 7.1

∴ PH = 14 - POH

PH = 14 - 7.1

= 6.9

2) and for Cu (OH) ^2

Ksp expression = [Cu+2][OH-]^2

by substitution:

2.2 x 10^-20 = 0.25 * [OH-]^2

∴[OH-] = 2.97 x 10^-10

POH = - ㏒[OH]

= - ㏒ (2.97 x 10^-10)

= 9.5

∴PH = 14 - POH

= 4.5

∴ Cu (OH) 2 will precipitate first because it needs less [OH-] and PH values than CO (OH) 2 to reach the Ksp value and reach equilibrium.

So you have to add more [OH-] than 2.97 x 10^-10 to precipitate Cu (OH) 2

and you have to add more [OH-] than 8 x 10^-8 to precipitate CO (OH) 2