Question

What is the minimum concentration of Cu2+ required to begin precipitating Cu(OH)2(s) in a solution of pH 10.77? For Cu(OH)2, Ksp= 2.6 x 10–19.

Answer 1

PH   = 10.77

POH   = 14-PH

          = 14-10.77

           = 3.23

-log[OH^-]   = 3.23

   [OH^-]   = 10^-3.23   = 0.000588M

Cu(OH)2(s) -----------------> Cu^2+ (aq) + 2OH^- (aq)

Ksp    = [Cu^2+][OH^-]^2

2.6*10^-19   = [Cu^2+]*(0.000588)^2

[Cu^2+]     = 2.6*10^-19/(0.000588)^2   = 7.52*10^-13M

[Cu^2+]       = 7.52*10^-13M >>>>>>>>>>>>>>answer