Question

The Ksp of yttrium iodate, Y(IO3)3 , is 1.12×10−10 . Calculate the molar solubility, s , of this compound.

Answer 1

Answer:

1.427x10^-3mol per L

Explanation:

$Y(IO_{3} )_{3} ---- Y^{3+} +IO_{3} ^{3-}$

I could use ⇌ in the math editor so I used ----

from the question each mole of Y(IO3)3 is dissolved  and this is giving us a mole of Y3+ and a mole of IO3^3-

Ksp = [Y^3+][IO3-]^3

So that,

1.12x10^-10 = [S][3S]^3

such that

1.12x10^-10 = 27S^4

the value of s is 0.001427mol per L

= 1.427x10^-3mol per L

so in conclusion

the molar solubility is therefore 1.427x10^-3mol per L