Diverse Ion effect and Activity coefficients problem

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SUMMARY

The discussion centers on calculating the solubility of Ba(IO3)2 in a 0.100M KIO3 solution at 25°C using activity coefficients. The Ksp value for Ba(IO3)2 is 1.57E-9. The user correctly identifies the ionic strengths and activity coefficients for Ba2+ (0.38) and IO3^- (0.775) but struggles with solving the cubic equation derived from the Ksp expression. The confusion arises from the presence of multiple solutions for X in the cubic equation, indicating a need for clarification on how to approach this problem.

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  • Understanding of solubility product constant (Ksp)
  • Familiarity with activity coefficients in ionic solutions
  • Knowledge of ICE (Initial, Change, Equilibrium) tables
  • Ability to solve cubic equations in chemistry
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  • Study the derivation and application of the solubility product constant (Ksp) in ionic compounds
  • Learn how to calculate activity coefficients using the Debye-Hückel equation
  • Practice solving cubic equations in chemical equilibrium contexts
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The question:

How many grams of Ba(IO3)2 can be dissolved in 700mL of a 0.100M KIO3 solution at 25C? use activity coefficients for this. (Ksp of Ba(IO3)2 =1.57E-9)

my answer:

I used this equation: Ksp = (concentration) * (activity coefficient)

well i found out that my ionic strength of KIO3 is 0.100M and from a chart, i found the respective activity coeff for Ba2+ and IO3-.
Ba2+ ---> 0.38
IO3^- ----> 0.775

Ba(IO3)2 <----> Ba2+ + 2IO3^-
I 0 x 0.100M
C +x +2x
E x 0.100M+2x


using the equation above,
Ksp = [Ba2+]*(activity coeff of Ba) * [IO3^-]^2 * (activity coeff of IO3^-)^2
1.57E-9 = (x)(0.38) * (0.100+2x)^2 * (0.775)^2

so i solve for X and that would be the concentration of Ba(IO3)2 in mol/L

BUT, to solve for X, i found it complicated because there would be 3 solutions for X aren't there?? since it's X^3... so i am confused on what to do.

help appreciated, Thanks.
 
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there above numbers below the balanced equation is supposed to be an ICE table
 

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