`500mL` of `2 xx 10^(-3)M A1C1_(3)` and `500mol` of `4 xx 10^(-2)M` solution of `NaOH` are mixed and solution is diluted to `10^(-2)L` with water at room temperature wil a precipiate exist? Given:
`K_(sp)`of `A1(OH)_(3) = 5 xx 10^(-33)`.

Explanation

`[A1^(3+)]:`
`{:(M_(1)V_(1),=,M_(2)V_(2),,),(("Before mixing"),,("After mixing"),,),(50xx2xx10^(-3),=,M_(2)xx1000,,):}`
(Total volume `= 50 + 500 = 1000 mL)`
`M_(2) = (500 xx 2xx 10^(-3))/(1000) = 10^(-3)M`
It is further diluted to `10^(2)L`, so `[A1^(3+)] = (10^(-3))/(10^(2)L) = 10^(-5)M`
Similarly `[overset(Theta)OH]: (4 xx 10^(-2))/(2xx10^(2)) = 2 xx 10^(-4)M`
`{:(A1(OH)_(3)rarr,2A1^(3+)+,3overset(Theta)OH,,),(,10^(-5)+,(2xx10^(-4)),,):}`
Hence, `Q_(sp) or I.P = [A1^(3+)] [overset(Theta)OH]^(3)`
`= 10^(-5) xx (2xx 10^(-4))^(3) = 8 xx 10^(-17)`.
`:. Q_(sp) gt K_(sp)`
Hence, precipitation of `A1(OH)_(3)` occurs in solution.