Question #885cc

1 Answer
il maestro
Apr 27, 2015

#[OH^-]# =#10^-5#

From #Ksp# of #Cu(OH)_2#, it starts precipitating at:
#2.2*10^-20# = #[Cu^(2+)]#*#[OH^-]^2#
#[OH^-]#=#sqrt (0.22*10^-20)#=#0.46*10^(-10)#

and from #Ksp# of #Mg(OH)_2#, it starts precipitating at
#6.3*10^-10# =#[Mg^(2+)]#*#[OH^-]^2#
#[OH^-]#=#sqrt (1.26*10^-10)# =#1.12*10^(-5)#

therefore at #[OH^-]#=#10^(-5)# all #Cu(OH)_2# is precipitated and all #Mg(OH)_2# is in solution

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