Calculate the molar solubility of BaSO4 in a solution in which H30 is (Ksp (BaSO4) = 1.1 x 10-10, K2(H2SO4) = = 1.02 x 10-2) %3D %3D а. 2.3 М. Molar solubility= M %3D b. 0.40 M. Molar solubility = M c. 0.057 M. Molar solubility = M %3D d. 0.16 M. Solubility = M %3D
Calculate the molar solubility of BaSO4 in a solution in which H30 is (Ksp (BaSO4) = 1.1 x 10-10, K2(H2SO4) = = 1.02 x 10-2) %3D %3D а. 2.3 М. Molar solubility= M %3D b. 0.40 M. Molar solubility = M c. 0.057 M. Molar solubility = M %3D d. 0.16 M. Solubility = M %3D
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Chapter1: Chemical Foundations
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![**Title:** Calculating the Molar Solubility of BaSO₄ in a H₃O⁺ Solution
**Instructions:**
Determine the molar solubility of barium sulfate (BaSO₄) in different concentrations of H₃O⁺.
**Given Values:**
- \( K_{sp} (\text{BaSO}_4) = 1.1 \times 10^{-10} \)
- \( K_2 (\text{H}_2\text{SO}_4) = 1.02 \times 10^{-2} \)
**Questions:**
a. For a solution where \([H_3O^+] = 2.3 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
b. For a solution where \([H_3O^+] = 0.40 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
c. For a solution where \([H_3O^+] = 0.057 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
d. For a solution where \([H_3O^+] = 0.16 \, \text{M}\).
Solubility = \(\_\_\_\_\_\_\) M
**Instructions for Completion:**
Calculate the molar solubility by considering the dissociation of BaSO₄ and its relationship with the \(K_{sp}\) and \(K_2\) values provided. Use these constants to find the equilibrium concentrations based on the given \( [H_3O^+]\) values.
**Note:** This exercise involves understanding the relationship between solubility product constants, dissociation constants, and their practical applications in solution chemistry.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fb2c2bfdd-c795-4c13-bc09-4b743040f0d9%2F1d0dc5be-cf79-402a-8dd7-d56acdf89598%2Fe7a5dx_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Title:** Calculating the Molar Solubility of BaSO₄ in a H₃O⁺ Solution
**Instructions:**
Determine the molar solubility of barium sulfate (BaSO₄) in different concentrations of H₃O⁺.
**Given Values:**
- \( K_{sp} (\text{BaSO}_4) = 1.1 \times 10^{-10} \)
- \( K_2 (\text{H}_2\text{SO}_4) = 1.02 \times 10^{-2} \)
**Questions:**
a. For a solution where \([H_3O^+] = 2.3 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
b. For a solution where \([H_3O^+] = 0.40 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
c. For a solution where \([H_3O^+] = 0.057 \, \text{M}\).
Molar solubility = \(\_\_\_\_\_\_\) M
d. For a solution where \([H_3O^+] = 0.16 \, \text{M}\).
Solubility = \(\_\_\_\_\_\_\) M
**Instructions for Completion:**
Calculate the molar solubility by considering the dissociation of BaSO₄ and its relationship with the \(K_{sp}\) and \(K_2\) values provided. Use these constants to find the equilibrium concentrations based on the given \( [H_3O^+]\) values.
**Note:** This exercise involves understanding the relationship between solubility product constants, dissociation constants, and their practical applications in solution chemistry.
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