If a is slowly added to a solution containing 0.0500 m of b and 0.0500 m of c, which solid will precipitate first? the solubility product constant for a2b3 is 2.30×10−8. the solubility product constant for ac4 is 4.10×10−8.

AC₄ will precipitate out first.

Explanation:

A solid will precipitate out if the ionic product of the solution exceeds the solubility product.

Let us check the ionic product

a) A₂B₃

Ionic product = [A]²[B]³

[A] = say "s"

[B] = 0.05 , [B]³ = (0.05)³ = 0.000125

2.3 X 10⁻⁸ = [A]²(0.000125)

[A] = 0.0136

b) AC₄

Ionic product = [A] [C]⁴

[A] = "s"

[A][0.05]⁴ = 4.10 X 10⁻⁸

[A]=0.00656 M

So for ionic product to exceed solubility product, we need less concentration of A in case of AC₄.

i think that the answer is c