Question

Chemistry 12 Question!!! [solubility]




The solubility of Mn(IO3)2 in water is 4.78x 10^-3 M. What [Mn2+] is required to just start precipitation of Mn(IO3)2 from a 0.0200 M solution of KIO3?

Please explain and show your steps to how to solve this question. Thank you.

Answer 1

First you have to find the Ksp, the solubility constant.

\[Ksp = [Mn][2IO3]^2\] - {because Mn(IO3)2} = Mn + 2IO3 }

As Mn(IO3)2 = 4.78*10^-3 /

Mn = 4.78*10^-3 and

2IO3 = 9.56*10^-3

(Note, when you have an equation that breaks down into X components and X does not equal 1, take it and multiply it by itself ergo: (x component) ^ x )

So!

\[Ksp = [4.78*10^{-3}][9.56*10^{-3}]^2 = 4.37*10^{-7}\]

Since you're putting Mn(IO3)2 in a solution which contains KIO3, you'll already have a few moles of IO3 in the solution. 0.02 moles to be exact.

Therefore since Ksp = 4.37*10^-7

and Ksp = [Mn][2IO3]^2

For this, just subtitute IO3 with 0.2 (the current number of moles of IO3 in the solution.) Ignore the coefficient, but do square it .

\[= Ksp = [Mn][0.02]^2\]

Divide Ksp/ (0.02)^2 and get the amount of Mn2+ required. (When broken off from IO3 it will become Mn+2 from Mn.)

\[4.37*10^{-7} / 0.02^2 = 1.09 *10^{-3} M\]