Answer:
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.
We conclude that in the case of AC_4 we need less of A
and hence AC_4 will be first to precipitate out.
a)Generally, equation for the ionic product is given as
Ionic product = [A]²[B]³
Where
[B] = 0.05
[B]³ = (0.05)³
[B] = 0.000125
Therefore
2.3 X 10^{-8} = [A]²(0.000125)
[A] = 0.0136
b)
Ionic product = [A] [C]⁴
[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.
Hence we conclude that in the case of AC_4 we need less of A
and hence AC_4 will [A]=0.00656 M
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