B.JAssuming that no equilibria other than dissolution are involved, calculate the molar solubility Peach of the following from its solubility product:
B.JAssuming that no equilibria other than dissolution are involved, calculate the molar solubility Peach of the following from its solubility product:
Chemistry
10th Edition
ISBN:9781305957404
Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste
Chapter1: Chemical Foundations
Section: Chapter Questions
Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...
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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+}][](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff7e0b628-1887-4036-8cfa-486903c53c1e%2Fccf716e3-f55c-4686-bec5-0c9ed8dd1987%2Fk7fu61n_processed.jpeg&w=3840&q=75)
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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