• korosh23
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.
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  • katieb
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  • Jadedry
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.) \[4.37*10^{-7} / 0.02^2 = 1.09 *10^{-3} M\]

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