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17.58 Calculate the solubility of LaF3\mathrm{LaF}_{3} in grams per liter in (a) ...
May 27, 2024
Solution
17.58 Calculate the solubility of $\mathrm{LaF}_{3}$ in grams per liter in:
1
Identify the solubility product constant ($K_{sp}$): The solubility product constant for LaF3\mathrm{LaF}_{3} is given by the expression Ksp=[La3+][F−]3K_{sp} = [\mathrm{La}^{3+}][\mathrm{F}^{-}]^3
2
Set up the solubility expression for pure water: Let ss be the molar solubility of LaF3\mathrm{LaF}_{3} in pure water. Then, [La3+]=s[\mathrm{La}^{3+}] = s and [F−]=3s[\mathrm{F}^{-}] = 3s. Therefore, Ksp=s⋅(3s)3=27s4K_{sp} = s \cdot (3s)^3 = 27s^4
3
Solve for $s$ in pure water: Given KspK_{sp} for LaF3\mathrm{LaF}_{3}, solve for ss using Ksp=27s4K_{sp} = 27s^4
4
Convert molar solubility to grams per liter: Multiply the molar solubility ss by the molar mass of LaF3\mathrm{LaF}_{3} to get the solubility in grams per liter
5
Consider the effect of common ions for KF solution: In a 0.010M0.010 \mathrm{M} KF solution, the concentration of F−\mathrm{F}^{-} ions is increased, which affects the solubility of LaF3\mathrm{LaF}_{3}. Use the modified KspK_{sp} expression to solve for the new solubility
6
Consider the effect of common ions for $\mathrm{LaCl}_{3}$ solution: In a 0.050MLaCl30.050 \mathrm{M} \mathrm{LaCl}_{3} solution, the concentration of La3+\mathrm{La}^{3+} ions is increased, which affects the solubility of LaF3\mathrm{LaF}_{3}. Use the modified KspK_{sp} expression to solve for the new solubility
Answer
The solubility of LaF3\mathrm{LaF}_{3} in pure water, 0.010M0.010 \mathrm{M} KF solution, and 0.050MLaCl30.050 \mathrm{M} \mathrm{LaCl}_{3} solution are calculated based on the KspK_{sp} and the common ion effect.
Key Concept
Solubility product constant (KspK_{sp}) and common ion effect
Explanation
The solubility of a sparingly soluble salt like LaF3\mathrm{LaF}_{3} is determined by its KspK_{sp} value. The presence of common ions in the solution affects the solubility by shifting the equilibrium.
17.62 Calculate the molar solubility of $\mathrm{Ni}(\mathrm{OH})_{2}$ when buffered at different pH levels:
1
Identify the solubility product constant ($K_{sp}$): The solubility product constant for Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} is given by the expression Ksp=[Ni2+][OH−]2K_{sp} = [\mathrm{Ni}^{2+}][\mathrm{OH}^{-}]^2
2
Set up the solubility expression for pH 8.0: At pH 8.0, [OH−]=10−(14−8)=10−6M[\mathrm{OH}^{-}] = 10^{-(14 - 8)} = 10^{-6} \mathrm{M}. Use this to solve for the molar solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2}
3
Set up the solubility expression for pH 10.0: At pH 10.0, [OH−]=10−(14−10)=10−4M[\mathrm{OH}^{-}] = 10^{-(14 - 10)} = 10^{-4} \mathrm{M}. Use this to solve for the molar solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2}
4
Set up the solubility expression for pH 12.0: At pH 12.0, [OH−]=10−(14−12)=10−2M[\mathrm{OH}^{-}] = 10^{-(14 - 12)} = 10^{-2} \mathrm{M}. Use this to solve for the molar solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2}
Answer
The molar solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} at pH 8.0, 10.0, and 12.0 are calculated based on the KspK_{sp} and the concentration of OH−\mathrm{OH}^{-} ions.
Key Concept
Solubility product constant (KspK_{sp}) and pH effect
Explanation
The solubility of Ni(OH)2\mathrm{Ni}(\mathrm{OH})_{2} is influenced by the pH of the solution, which affects the concentration of OH−\mathrm{OH}^{-} ions.
17.70 (a) Will $\mathrm{Co}(\mathrm{OH})_{2}$ precipitate from solution if the pH of a $0.020 \mathrm{M}$ solution of $\mathrm{Co}\left(\mathrm{NO}_{3}\right)_{2}$ is adjusted to 8.5?
1
Identify the solubility product constant ($K_{sp}$): The solubility product constant for Co(OH)2\mathrm{Co}(\mathrm{OH})_{2} is given by the expression Ksp=[Co2+][OH−]2K_{sp} = [\mathrm{Co}^{2+}][\mathrm{OH}^{-}]^2
2
Calculate $[\mathrm{OH}^{-}]$ at pH 8.5: At pH 8.5, [OH−]=10−(14−8.5)=10−5.5M[\mathrm{OH}^{-}] = 10^{-(14 - 8.5)} = 10^{-5.5} \mathrm{M}
3
Determine the ion product (Q): Calculate the ion product Q=[Co2+][OH−]2Q = [\mathrm{Co}^{2+}][\mathrm{OH}^{-}]^2 using [Co2+]=0.020M[\mathrm{Co}^{2+}] = 0.020 \mathrm{M} and [OH−]=10−5.5M[\mathrm{OH}^{-}] = 10^{-5.5} \mathrm{M}
4
Compare $Q$ with $K_{sp}$: If Q > K_{sp}, precipitation occurs. If Q < K_{sp}, no precipitation occurs
Answer
Whether Co(OH)2\mathrm{Co}(\mathrm{OH})_{2} will precipitate at pH 8.5 is determined by comparing the ion product QQ with the KspK_{sp}.
Key Concept
Solubility product constant (KspK_{sp}) and ion product (Q)
Explanation
Precipitation occurs when the ion product QQ exceeds the solubility product constant KspK_{sp}.
17.70 (b) Will $\mathrm{AgIO}_{3}$ precipitate when $20 \mathrm{~mL}$ of $0.010 \mathrm{M} \mathrm{AgIO}_{3}$ is mixed with $10 \mathrm{~mL}$ of $0.015 \mathrm{M} \mathrm{NaIO}_{3}$?
1
Identify the solubility product constant ($K_{sp}$): The solubility product constant for AgIO3\mathrm{AgIO}_{3} is Ksp=3.1×10−8K_{sp} = 3.1 \times 10^{-8}
2
Calculate the final concentrations: Use the dilution formula to find the final concentrations of Ag+\mathrm{Ag}^{+} and IO3−\mathrm{IO}_{3}^{-} ions after mixing
3
Determine the ion product (Q): Calculate the ion product Q=[Ag+][IO3−]Q = [\mathrm{Ag}^{+}][\mathrm{IO}_{3}^{-}] using the final concentrations
4
Compare $Q$ with $K_{sp}$: If Q > K_{sp}, precipitation occurs. If Q < K_{sp}, no precipitation occurs
Answer
Whether AgIO3\mathrm{AgIO}_{3} will precipitate when mixed is determined by comparing the ion product QQ with the KspK_{sp}.
Key Concept
Solubility product constant (KspK_{sp}) and ion product (Q)
Explanation
Precipitation occurs when the ion product QQ exceeds the solubility product constant KspK_{sp}.
17.74 A solution of $\mathrm{Na}_{2} \mathrm{SO}_{4}$ is added dropwise to a solution that is $0.010 \mathrm{M}$ in $\mathrm{Ba}^{2+}(a q)$ and $0.010 \mathrm{M}$ in $\mathrm{Sr}^{2+}(a q)$.
1
Identify the solubility product constants ($K_{sp}$): The solubility product constants are Ksp(BaSO4)=1.1×10−10K_{sp}(\mathrm{BaSO}_{4}) = 1.1 \times 10^{-10} and Ksp(SrSO4)=3.2×10−7K_{sp}(\mathrm{SrSO}_{4}) = 3.2 \times 10^{-7}
2
Calculate the concentration of $\mathrm{SO}_{4}^{2-}$ for $\mathrm{BaSO}_{4}$ precipitation: Use Ksp(BaSO4)=[Ba2+][SO42−]K_{sp}(\mathrm{BaSO}_{4}) = [\mathrm{Ba}^{2+}][\mathrm{SO}_{4}^{2-}] to find the concentration of SO42−\mathrm{SO}_{4}^{2-} needed to begin precipitation
3
Calculate the concentration of $\mathrm{SO}_{4}^{2-}$ for $\mathrm{SrSO}_{4}$ precipitation: Use Ksp(SrSO4)=[Sr2+][SO42−]K_{sp}(\mathrm{SrSO}_{4}) = [\mathrm{Sr}^{2+}][\mathrm{SO}_{4}^{2-}] to find the concentration of SO42−\mathrm{SO}_{4}^{2-} needed to begin precipitation
4
Determine which cation precipitates first: Compare the concentrations of SO42−\mathrm{SO}_{4}^{2-} needed for BaSO4\mathrm{BaSO}_{4} and SrSO4\mathrm{SrSO}_{4} precipitation to determine which cation precipitates first
5
Calculate the concentration of $\mathrm{SO}_{4}^{2-}$ when the second cation begins to precipitate: Use the concentration of SO42−\mathrm{SO}_{4}^{2-} when the first cation has precipitated to find the concentration when the second cation begins to precipitate
Answer
The concentration of SO42−\mathrm{SO}_{4}^{2-} necessary to begin precipitation, the cation that precipitates first, and the concentration of SO42−\mathrm{SO}_{4}^{2-} when the second cation begins to precipitate are determined based on the KspK_{sp} values.
Key Concept
Solubility product constant (KspK_{sp}) and selective precipitation
Explanation
The order of precipitation and the concentration of ions required for precipitation are determined by the solubility product constants (KspK_{sp}) of the salts involved.
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