**Calculating Adjusted Solubility Product Constants, $ K'_{sp} $**For a solution in which the ionic strength $\mu = 7.5 \times 10^{-2}$, calculate the adjusted solubility product $ K'_{sp} $ for the following compounds:a. **AgSCN** - Given: $ K_{sp}(\text{AgSCN}) = 1.1 \times 10^{-12} $ - Activity coefficients: $ \alpha_{\text{Ag}^+} = 0.25 $, $ \alpha_{\text{SCN}^-} = 0.35 $ - $ K'_{sp} = $ [Blank for calculation]b. **PbI$_2$** - Given: $ K_{sp}(\text{PbI}_2) = 7.9 \times 10^{-9} $ - Activity coefficients: $ \alpha_{\text{Pb}^{2+}} = 0.45 $, $ \alpha_{\text{I}^-} = 0.3 $ - $ K'_{sp} = $ [Blank for calculation]c. **La(IO$_3$)$_3$** - Given: $ K_{sp}(\text{La(IO}_3)_3) = 1.0 \times 10^{-11} $ - Activity coefficients: $ \alpha_{\text{La}^{3+}} = 0.9 $, $ \alpha_{\text{IO}_3^-} = 0.35 $ - $ K'_{sp} = $ [Blank for calculation]**Instructions:**To find the adjusted solubility product $ K'_{sp} $, use the activity coefficients provided for each ion in the solution. The formulas for each compound entail adjusting the original $ K_{sp} $ value with the activity coefficients as follows:\[ K'_{sp} = K_{sp} \times (\alpha_{\text{cation}}) \times (\alpha_{\text{anion}}^{\text{stoichiometric coefficient}}) \]Make sure to substitute the given values into the formula and solve for $ K'_{sp} $.

a. Ksp of AgSCN = 1.1 x 10-12 , b. Ksp of PbI2 = 7.9 x 10-9 , c. Ksp of La(IO3)3 = 1.0 x 10-11 ,

Answered: For a solution in which µ = 7.5 × 10-²,…

Chemistry

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ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

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**Calculating Adjusted Solubility Product Constants, $ K'_{sp} $**

For a solution in which the ionic strength $\mu = 7.5 \times 10^{-2}$, calculate the adjusted solubility product $ K'_{sp} $ for the following compounds:

a. **AgSCN**
- Given: $ K_{sp}(\text{AgSCN}) = 1.1 \times 10^{-12} $
- Activity coefficients: $ \alpha_{\text{Ag}^+} = 0.25 $, $ \alpha_{\text{SCN}^-} = 0.35 $
- $ K'_{sp} = $ [Blank for calculation]

b. **PbI$_2$**
- Given: $ K_{sp}(\text{PbI}_2) = 7.9 \times 10^{-9} $
- Activity coefficients: $ \alpha_{\text{Pb}^{2+}} = 0.45 $, $ \alpha_{\text{I}^-} = 0.3 $
- $ K'_{sp} = $ [Blank for calculation]

c. **La(IO$_3$)$_3$**
- Given: $ K_{sp}(\text{La(IO}_3)_3) = 1.0 \times 10^{-11} $
- Activity coefficients: $ \alpha_{\text{La}^{3+}} = 0.9 $, $ \alpha_{\text{IO}_3^-} = 0.35 $
- $ K'_{sp} = $ [Blank for calculation]

**Instructions:**

To find the adjusted solubility product $ K'_{sp} $, use the activity coefficients provided for each ion in the solution. The formulas for each compound entail adjusting the original $ K_{sp} $ value with the activity coefficients as follows:

\[ K'_{sp} = K_{sp} \times (\alpha_{\text{cation}}) \times (\alpha_{\text{anion}}^{\text{stoichiometric coefficient}}) \]

Make sure to substitute the given values into the formula and solve for $ K'_{sp} $.

Expert Solution

Step 1: Given data of the question

a. Ksp of AgSCN = 1.1 x 10^-12 , $alpha subscript A g to the power of plus space end subscript equals space 0.25 space comma space alpha subscript S C N to the power of minus space end subscript equals space 0.35$

b. Ksp of PbI₂ = 7.9 x 10^-9, $alpha subscript P b to the power of 2 plus end exponent space end subscript equals space 0.45 space comma space alpha subscript I to the power of minus space end subscript equals space 0.3$

c. Ksp of La(IO₃)₃ = 1.0 x 10^-11 , $alpha subscript L a to the power of 3 plus end exponent space end subscript equals space 0.9 space comma space alpha subscript I O subscript 3 to the power of minus space end subscript equals space 0.35$

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