Question

The Ksp of Al(OH)3 at 25 oC is 1 x 10-33. Consider a solution that is 1.0 x 10-10 M Al(NO3)3 and 1.0 x 10-7 M NaOH. Selected Answer: C. Q > Ksp and a precipitate will form. Correct Answer: C. Q > Ksp and a precipitate will form.

Answer 1

Answer :
`Q>K_(sp)` and a precipitate will form.

Explanation : Given,

Concentration of
`Al(NO_3)_3` =
`1.0* 10^(-10)M`

Concentration of
`NaOH` =
`1.0* 10^(-7)M`

`K_(sp)` of
`Al(OH)_3` =
`1* 10^(-33)`

The equilibrium chemical reaction will be:

`Al(OH)_3\rightleftharpoons Al^(3+)+3OH^-`

The expression of Q for this reaction is:

`Q=[Al^(3+)][OH^-]^3`

`Al(NO_3)_3\rightleftharpoons Al^(3+)+3NO_3^-`

Concentration of
`Al(NO_3)_3` = Concentration of
`Al^(3+)` =
`1.0* 10^(-10)M`

`NaOH\rightleftharpoons Na^(+)+OH^-`

Concentration of
`NaOH` = Concentration of
`OH^-` =
`1.0* 10^(-7)M`

Now put all the given values in this expression, we get:

`Q=(1.0* 10^(-10))* (1.0* 10^(-7))^3`

`Q=1.0* 10^(-31)`

There are 3 conditions:

When
`K_(sp)>Q_(sp)`; the reaction is product favored. (No precipitation)

When
`K_(sp)<Q_(sp)`; the reaction is reactant favored. (Precipitation)

When
`K_(sp)=Q_(sp)`; the reaction is in equilibrium. (Sparingly soluble)

As, the
`Q_(sp)` is more than
`K_(sp)`. The above reaction is reactant favored. This means salt or precipitate will be formed.

Hence, the
`Q>K_(sp)` and a precipitate will form.