Chemistry
Chemistry
10th Edition
ISBN: 9781305957404
Author: Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher: Cengage Learning
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3x10^-4,1.4x10^-8,1.55x10^-41,4.5x10^-29 how did we figure these number out????I am so confused??
**Calculating Molar Solubility from Solubility Product**

**Problem Statement:**

13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product:
(a) KHC₄H₄O₆ 
(b) PbI₂
(c) Ag₄[Fe(CN)₆], a salt containing the \( Fe(CN)_6^{4-} \) ion
(d) Hg₂I₂

**Solution:**

Let \( x \) be the molar solubility.

(a) For KHC₄H₄O₆:
\[ K_{sp} = [K^+][HC₄H₄O₆^-] = 3 \times 10^{-4} \]
\[ x^2 = 3 \times 10^{-4} \]
\[ x = \sqrt{3 \times 10^{-4}} = 2 \times 10^{-2} M \]

(b) For PbI₂:
\[ K_{sp} = [Pb^{2+}][I^-]^2 = 1.4 \times 10^{-8} \]
\[ x[2x]^2 = 1.4 \times 10^{-8} \]
\[ 4x^3 = 1.4 \times 10^{-8} \]
\[ x = \left(\frac{1.4 \times 10^{-8}}{4}\right)^{\frac{1}{3}} = 1.5 \times 10^{-3} M \]

(c) For Ag₄[Fe(CN)₆]:
\[ K_{sp} = [Ag^+]^4[Fe(CN)_6^{4-}] = 1.55 \times 10^{-41} \]
\[ (4x)^4 x = 1.55 \times 10^{-41} \]
\[ 256x^5 = 1.55 \times 10^{-41} \]
\[ x = \left(\frac{1.55 \times 10^{-41}}{256}\right)^{\frac{1}{5}} = 2.27 \times 10^{-9} M \]

(d) For Hg₂I₂:
\[ K_{sp} = [Hg₂^{2+}][
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Transcribed Image Text:**Calculating Molar Solubility from Solubility Product** **Problem Statement:** 13. Assuming that no equilibria other than dissolution are involved, calculate the molar solubility of each of the following from its solubility product: (a) KHC₄H₄O₆ (b) PbI₂ (c) Ag₄[Fe(CN)₆], a salt containing the \( Fe(CN)_6^{4-} \) ion (d) Hg₂I₂ **Solution:** Let \( x \) be the molar solubility. (a) For KHC₄H₄O₆: \[ K_{sp} = [K^+][HC₄H₄O₆^-] = 3 \times 10^{-4} \] \[ x^2 = 3 \times 10^{-4} \] \[ x = \sqrt{3 \times 10^{-4}} = 2 \times 10^{-2} M \] (b) For PbI₂: \[ K_{sp} = [Pb^{2+}][I^-]^2 = 1.4 \times 10^{-8} \] \[ x[2x]^2 = 1.4 \times 10^{-8} \] \[ 4x^3 = 1.4 \times 10^{-8} \] \[ x = \left(\frac{1.4 \times 10^{-8}}{4}\right)^{\frac{1}{3}} = 1.5 \times 10^{-3} M \] (c) For Ag₄[Fe(CN)₆]: \[ K_{sp} = [Ag^+]^4[Fe(CN)_6^{4-}] = 1.55 \times 10^{-41} \] \[ (4x)^4 x = 1.55 \times 10^{-41} \] \[ 256x^5 = 1.55 \times 10^{-41} \] \[ x = \left(\frac{1.55 \times 10^{-41}}{256}\right)^{\frac{1}{5}} = 2.27 \times 10^{-9} M \] (d) For Hg₂I₂: \[ K_{sp} = [Hg₂^{2+}][
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