The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated: a. AgBr:[Ag*] = 5.7 × 10-7 M, [Br]= 5.7 × 10-7M b. CaCO3: [Ca²+] = 5.3 × 10-³M, [CO3²¯] = 9.0 × 10-7M c. PbF₂: [Pb²+] = 2.1 × 10-³ M, [F] = 4.2 × 10-³ M
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate Ksp for each of the slightly soluble solids indicated: a. AgBr:[Ag*] = 5.7 × 10-7 M, [Br]= 5.7 × 10-7M b. CaCO3: [Ca²+] = 5.3 × 10-³M, [CO3²¯] = 9.0 × 10-7M c. PbF₂: [Pb²+] = 2.1 × 10-³ M, [F] = 4.2 × 10-³ M
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![### Calculating the Solubility Product Constant (Ksp)
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \( K_{sp} \) for each of the slightly soluble solids indicated:
a. **Silver Bromide (AgBr):**
\([\text{Ag}^+]\) = \(5.7 \times 10^{-7}\) M, \([\text{Br}^-]\) = \(5.7 \times 10^{-7}\) M
b. **Calcium Carbonate (CaCO₃):**
\([\text{Ca}^{2+}]\) = \(5.3 \times 10^{-3}\) M, \([\text{CO}_3^{2-}]\) = \(9.0 \times 10^{-7}\) M
c. **Lead(II) Fluoride (PbF₂):**
\([\text{Pb}^{2+}]\) = \(2.1 \times 10^{-3}\) M, \([\text{F}^-]\) = \(4.2 \times 10^{-3}\) M
#### Explanation:
In order to calculate the solubility product constant (\( K_{sp} \)), we use the concentrations of the ions at equilibrium. The general formula for the solubility product constant is:
\[ K_{sp} = [\text{Cation}^{n+}] [\text{Anion}^{m-}] \]
Where \([\text{Cation}^{n+}]\) and \([\text{Anion}^{m-}]\) are the molar concentrations of the ions in solution.
For the given compounds:
**a. AgBr**
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
**b. CaCO₃**
\[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \]
**c. PbF₂**
\[ K_{sp} = [\text{Pb}^{2+}][\text{F}^-]^2 \]
Note that for PbF₂, the fluoride ion concentration is squared because each formula unit of PbF₂ produces two fluoride ions.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Ff3c22a28-8e9a-42a1-ac26-114f2e495bee%2Fe0e47ca3-2bf9-42a6-979e-94c59fadfd0d%2Fkjwgk3o_processed.png&w=3840&q=75)
Transcribed Image Text:### Calculating the Solubility Product Constant (Ksp)
The following concentrations are found in mixtures of ions in equilibrium with slightly soluble solids. From the concentrations given, calculate \( K_{sp} \) for each of the slightly soluble solids indicated:
a. **Silver Bromide (AgBr):**
\([\text{Ag}^+]\) = \(5.7 \times 10^{-7}\) M, \([\text{Br}^-]\) = \(5.7 \times 10^{-7}\) M
b. **Calcium Carbonate (CaCO₃):**
\([\text{Ca}^{2+}]\) = \(5.3 \times 10^{-3}\) M, \([\text{CO}_3^{2-}]\) = \(9.0 \times 10^{-7}\) M
c. **Lead(II) Fluoride (PbF₂):**
\([\text{Pb}^{2+}]\) = \(2.1 \times 10^{-3}\) M, \([\text{F}^-]\) = \(4.2 \times 10^{-3}\) M
#### Explanation:
In order to calculate the solubility product constant (\( K_{sp} \)), we use the concentrations of the ions at equilibrium. The general formula for the solubility product constant is:
\[ K_{sp} = [\text{Cation}^{n+}] [\text{Anion}^{m-}] \]
Where \([\text{Cation}^{n+}]\) and \([\text{Anion}^{m-}]\) are the molar concentrations of the ions in solution.
For the given compounds:
**a. AgBr**
\[ K_{sp} = [\text{Ag}^+][\text{Br}^-] \]
**b. CaCO₃**
\[ K_{sp} = [\text{Ca}^{2+}][\text{CO}_3^{2-}] \]
**c. PbF₂**
\[ K_{sp} = [\text{Pb}^{2+}][\text{F}^-]^2 \]
Note that for PbF₂, the fluoride ion concentration is squared because each formula unit of PbF₂ produces two fluoride ions.
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