A saturated solution of #Mg(OH)_2# in water has pH = 10.32. How do you calculate the Ksp of #Mg(OH)_2#?

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Answer 1 · 2016-06-15T16:02:07

You can do it like this:

Magnesium hydroxide is sparingly soluble and in solution the solid is in equilibrium with the aqueous ions:

#Mg(OH)_(2)rightleftharpoonsMg^(2+)+2OH^(-)#

We are told the #pH# so we can find #[OH^-]#.

#pH+pOH=14#

#:.pOH=14-pH=14-10.3=3.7#

#:.-log[OH^-]=3.7#

From which #[OH^-]=2xx10^(-4)" ""mol/l"#

From the equation you can see that #[Mg^(2+)]# must be half of this.

#:.[Mg^(2+)]=(2xx10^(-4))/2=10^(-4)" ""mol/l"#

The expression for #K_(sp)# is given by:

#K_(sp)=[Mg^(2+)][OH^(-)]^2#

Putting in the numbers:

#K_(sp)=10^(-4)xx(2xx10^(-4))^2=4xx10^(-12)" ""mol"^(3)."l"^(-3)#