Solid sodium sulfate, NazSO4, is added slowly to a solution that is 1.0 x 104 in both Ba2+ and Pb²+ until [SO4²-] reaches 1.0 x 104 M. Would either BaSO4 (Kan = 1.1 x 10-10) or P6SO4 (Kan = 1.7 x 10-8) precipitate under these conditions? %3D
Solid sodium sulfate, NazSO4, is added slowly to a solution that is 1.0 x 104 in both Ba2+ and Pb²+ until [SO4²-] reaches 1.0 x 104 M. Would either BaSO4 (Kan = 1.1 x 10-10) or P6SO4 (Kan = 1.7 x 10-8) precipitate under these conditions? %3D
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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![**Question:**
Solid sodium sulfate, \( \text{Na}_2\text{SO}_4 \), is added slowly to a solution that is \( 1.0 \times 10^{-4} \) M in both \( \text{Ba}^{2+} \) and \( \text{Pb}^{2+} \) until \([ \text{SO}_4^{2-} ]\) reaches \( 1.0 \times 10^{-4} \) M. Would either \( \text{BaSO}_4 \) (\( K_{sp} = 1.1 \times 10^{-10} \)) or \( \text{PbSO}_4 \) (\( K_{sp} = 1.7 \times 10^{-8} \)) precipitate under these conditions?
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**Explanation:**
This problem requires you to determine if either barium sulfate (\( \text{BaSO}_4 \)) or lead(II) sulfate (\( \text{PbSO}_4 \)) will precipitate when the concentration of sulfate ions (\( \text{SO}_4^{2-} \)) reaches \( 1.0 \times 10^{-4} \) M in the given solution.
#### Key Information:
- Initial concentrations: \( [\text{Ba}^{2+}] = 1.0 \times 10^{-4} \) M, \( [\text{Pb}^{2+}] = 1.0 \times 10^{-4} \) M
- \( K_{sp} \) for \( \text{BaSO}_4 = 1.1 \times 10^{-10} \)
- \( K_{sp} \) for \( \text{PbSO}_4 = 1.7 \times 10^{-8} \)
#### Approach:
1. Calculate the ion product (Q) for both \( \text{BaSO}_4 \) and \( \text{PbSO}_4 \).
- For \( \text{BaSO}_4 \): \( Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \)
- For \( \text{PbSO}_4 \): \( Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \)
2. Compare the ion product](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F24666b77-75da-49bd-bf84-8c4674dedd2c%2Fd3ae93aa-c497-4895-892e-7045e09484c3%2Fc93ppkm.png&w=3840&q=75)
Transcribed Image Text:**Question:**
Solid sodium sulfate, \( \text{Na}_2\text{SO}_4 \), is added slowly to a solution that is \( 1.0 \times 10^{-4} \) M in both \( \text{Ba}^{2+} \) and \( \text{Pb}^{2+} \) until \([ \text{SO}_4^{2-} ]\) reaches \( 1.0 \times 10^{-4} \) M. Would either \( \text{BaSO}_4 \) (\( K_{sp} = 1.1 \times 10^{-10} \)) or \( \text{PbSO}_4 \) (\( K_{sp} = 1.7 \times 10^{-8} \)) precipitate under these conditions?
---
**Explanation:**
This problem requires you to determine if either barium sulfate (\( \text{BaSO}_4 \)) or lead(II) sulfate (\( \text{PbSO}_4 \)) will precipitate when the concentration of sulfate ions (\( \text{SO}_4^{2-} \)) reaches \( 1.0 \times 10^{-4} \) M in the given solution.
#### Key Information:
- Initial concentrations: \( [\text{Ba}^{2+}] = 1.0 \times 10^{-4} \) M, \( [\text{Pb}^{2+}] = 1.0 \times 10^{-4} \) M
- \( K_{sp} \) for \( \text{BaSO}_4 = 1.1 \times 10^{-10} \)
- \( K_{sp} \) for \( \text{PbSO}_4 = 1.7 \times 10^{-8} \)
#### Approach:
1. Calculate the ion product (Q) for both \( \text{BaSO}_4 \) and \( \text{PbSO}_4 \).
- For \( \text{BaSO}_4 \): \( Q = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \)
- For \( \text{PbSO}_4 \): \( Q = [\text{Pb}^{2+}][\text{SO}_4^{2-}] \)
2. Compare the ion product
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