In an aqueous solution containing #"1.0 M"# #"NH"_4"Cl"# (#K_a = 5.56 xx 10^(-10)#), what is the solubility of #"Mg"("OH")_2#? #K_(sp) = 5.5 xx 10^(-11)#.

the solubility of #Mg(OH)_2# is 0,014 g/L that divided for its MM =58g/mol give a molar solubility of #2,4 xx 10^-4 (mol)/L#
its ionic dissociation give

#Mg(OH)_2 = Mg^2+ +2OH^-#

hence the amount of #OH^-# will be #4,8 xx 10^-4 (mol)/L# the costant of dissociation of #Mg(OH)_2# is:

#K_S=[Mg^(2+)] xx [OH^-]^2= 5,5 xx 10^(-11)#

As #Mg(OH)_2# is a base weaker of ammonia it doesn't react with #NH_4Cl#

#NH_4Cl# whose pH is
#[H^+]= sqrt (K_w/K_b xx Cs)=sqrt(10^(-14)/(1,8 xx10^(-5))xx 1= 2,3 xx 10^-5# that is the amount of the #[H^+]# and give the acidity of the solution
(pH=4,64) in the solution and the amount of [OH^-] wil be:
#[OH^-]= 10^(-14)/(2,3 xx 10^-5)=4,3 xx 10^(-10)#
now if in the costant of dissociation of #Mg(OH)_2#:
#K_S=[Mg^(2+)] xx [OH^-]^2= 5,5 xx 10^(-11)#
you put the OH^- of the #NH_4Cl# solution in this expression you obtain for #[Mg^(2+)]#a very high value that minds that #Mg(OH)_2# is completely soluble at pH 4,64.

But I think that the solubiliy dipends from the amount of the two substances

I got #2.48 xx 10^(-4) "M"#, a mere #8 xx 10^(-6) "M"# more soluble than in pure water.

The #K_(sp)# is #"Mg"("OH")_2# is #5.5 xx 10^(-11)#, and the #K_a# of #"NH"_4^(+)# is #5.56 xx 10^(-10)#.

What we should recognize is that the #"H"^(+)# generated from the #"NH"_4^(+)# dissociation reacts with the #"OH"^(-)# from #"Mg"("OH")_2# dissolution. So the solubility should increase, at least a little...

First, we let the #"NH"_4"Cl"# dissociate to see how much #"H"^(+)# we get. The ICE table in #"M"# is:

#"NH"_4^(+)(aq) rightleftharpoons "NH"_3(aq) + "H"^(+)(aq)#

#"I"" "1.0" "" "" "" "0" "" "" "" "0#
#"C"" "-x" "" "" "+x" "" "" "+x#
#"E"" "1.0-x" "" "x" "" "" "" "x#

Therefore,

#5.56 xx 10^(-10) = x^2/(1.0 - x)#

This is obviously a small #K_a#, so we can make the small #x# approximation, i.e. #x# #"<<"# #"1.0 M"#. As a result,

#5.56 xx 10^(-10) ~~ x^2/1.0#

#=> x = ["H"^(+)] ~~ 2.36 xx 10^(-5) "M"#

(And indeed it is much smaller than #"1.0 M"#.)

Now, this amount of #"H"^(+)# can then neutralize whatever #"OH"^(-)# is in solution from #"Mg"("OH")_2# dissolution in pure water...

#K_(sp) = 5.5 xx 10^(-11) = ["Mg"^(2+)]["OH"^(-)]^2#

#= (s)(2s)^2 = 4s^3#

#=> s = (K_(sp)/4)^(1//3) = 2.40 xx 10^(-4) "M Mg"^(2+)#

Or, we get #["OH"^(-)] = 2s = 4.80 xx 10^(-4) "M"#. The #"H"^(+)# then neutralizes some of this, reducing the #Q_(sp)# so that #Q_(sp) < K_(sp)#.

Therefore, #Q_(sp)# wants to increase by making more #"OH"^(-)# by Le Chatelier's principle, in turn making more #"Mg"^(2+)# as well.

#K_(sp) = 5.5 xx 10^(-11)#

#= (2.40 xx 10^(-4) + s')(overbrace(4.80 xx 10^(-4))^("starting OH"^(-)) - overbrace(2.36 xx 10^(-5))^("lost OH"^(-)) + overbrace(2s')^("Le Chatelier gain in OH"^(-)))^2#

#5.5 xx 10^(-11) = (2.40 xx 10^(-4) + s')(4.56 xx 10^(-4) + 2s')^2#

This becomes the cubic

#4s'^3 + 2.78 xx 10^(-3)s'^2 + 6.46 xx 10^(-7) s' + (4.99 - 5.5) xx 10^(-11) = 0#

You could not make the approximation that #(4.99 - 5.5) xx 10^(-11) ~~ 0#, as it would give imaginary concentrations.

So, this actually gives #s'# not small, and there is no good approximation that can be made here that I know of. Upon solving the cubic numerically, the result is #s' = 7.637 xx 10^(-6) "M"#.

Therefore, #["Mg"^(2+)]_(eq) = "molar solubility"#, which is

#color(blue)(["Mg"("OH")_2(aq)]) = ["Mg"^(+)]_(eq)#

#= 2.40 xx 10^(-4) + s' ~~ color(blue)(2.48 xx 10^(-4) "M")#

Not much of an improvement over its normal solubility of #2.40 xx 10^(-4) "M"#! That's what a weak acid can do, apparently.