Respuesta :
Answer : [tex]Q>K_{sp}[/tex] and a precipitate will form.
Explanation : Given,
Concentration of [tex]Al(NO_3)_3[/tex] = [tex]1.0\times 10^{-10}M[/tex]
Concentration of [tex]NaOH[/tex] = [tex]1.0\times 10^{-7}M[/tex]
[tex]K_{sp}[/tex] of [tex]Al(OH)_3[/tex] = [tex]1\times 10^{-33}[/tex]
The equilibrium chemical reaction will be:
[tex]Al(OH)_3\rightleftharpoons Al^{3+}+3OH^-[/tex]
The expression of Q for this reaction is:
[tex]Q=[Al^{3+}][OH^-]^3[/tex]
[tex]Al(NO_3)_3\rightleftharpoons Al^{3+}+3NO_3^-[/tex]
Concentration of [tex]Al(NO_3)_3[/tex] = Concentration of [tex]Al^{3+}[/tex] = [tex]1.0\times 10^{-10}M[/tex]
[tex]NaOH\rightleftharpoons Na^{+}+OH^-[/tex]
Concentration of [tex]NaOH[/tex] = Concentration of [tex]OH^-[/tex] = [tex]1.0\times 10^{-7}M[/tex]
Now put all the given values in this expression, we get:
[tex]Q=(1.0\times 10^{-10})\times (1.0\times 10^{-7})^3[/tex]
[tex]Q=1.0\times 10^{-31}[/tex]
There are 3 conditions:
When [tex]K_{sp}>Q_{sp}[/tex]; the reaction is product favored. (No precipitation)
When [tex]K_{sp}<Q_{sp}[/tex]; the reaction is reactant favored. (Precipitation)
When [tex]K_{sp}=Q_{sp}[/tex]; the reaction is in equilibrium. (Sparingly soluble)
As, the [tex]Q_{sp}[/tex] is more than [tex]K_{sp}[/tex]. The above reaction is reactant favored. This means salt or precipitate will be formed.
Hence, the [tex]Q>K_{sp}[/tex] and a precipitate will form.
