Problem 74
Question
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 \(\mathrm{Sr}^{2+}(a q) .(\mathbf{a}) \mathrm{What}\) concentration of \(\mathrm{SO}_{4}^{2-}\) is necessary to begin precipitation? (Neglect volume changes. \(\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10} ; \mathrm{SrSO}_{4}:\) \(K_{s p}=3.2 \times 10^{-7} .\) ) (b) Which cation precipitates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}(a q)\) when the second cation begins to precipitate?
Step-by-Step Solution
Verified Answer
(a) The necessary concentration of $\mathrm{SO}_{4}^{2-}$ to begin precipitation is \(1.1\times10^{-9}\,M\).
(b) $\mathrm{Ba}^{2+}$ will precipitate first.
(c) The concentration of $\mathrm{SO}_{4}^{2-}$ needed for the second cation, $\mathrm{Sr}^{2+}$, to precipitate is \(3.2\times10^{-6}\,M\).
1Step 1: Write solubility product expressions.
The solubility products expressions can be written as:
For BaSO4:
\[K_{sp}(BaSO_4) = [Ba^{2+}][SO_4^{2-}]\]
For SrSO4:
\[K_{sp}(SrSO_4) = [Sr^{2+}][SO_4^{2-}]\]
2Step 2: Find saturation concentrations for each cation.
The concentration of Ba^2+ and Sr^2+ ions are both given as 0.010 M.
Now, let's solve for the saturation concentration of SO4^2- for each cation using the solubility products:
For Ba^2+:
\[[SO_4^{2-}]_{sat}(Ba^{2+}) = \frac{K_{sp}(BaSO_4)}{[Ba^{2+}]} = \frac{1.1\times10^{-10}}{0.010} = 1.1\times10^{-9}\,M\]
For Sr^2+:
\[[SO_4^{2-}]_{sat}(Sr^{2+}) = \frac{K_{sp}(SrSO_4)}{[Sr^{2+}]} = \frac{3.2\times10^{-7}}{0.010} = 3.2\times10^{-6}\,M\]
3Step 3: Determine which cation precipitates first.
Since we are looking for the minimum concentration of SO4^2- to start precipitation, we should compare the saturation concentrations calculated in step 2. The cation with the lower saturation concentration will precipitate first:
\(1.1\times10^{-9}\,M < 3.2\times10^{-6}\,M\), therefore Ba^2+ will precipitate first.
Answer (b): Ba^2+ will precipitate first.
4Step 4: Calculate the concentration of SO4^2- when the second cation begins to precipitate.
After Ba^2+ has precipitated, it is now out of the solution and does not affect the solution anymore. The concentration of SO4^2- required to make Sr^2+ precipitate can be found using the saturation concentration we calculated for Sr^2+:
Answer (c): The concentration of SO4^2- when the second cation (Sr^2+) begins to precipitate is \(3.2\times10^{-6}\,M\).
In summary:
(a) The necessary concentration of sulfate ions to begin precipitation is \(1.1\times10^{-9}\,M\).
(b) Ba^2+ will precipitate first.
(c) The concentration of sulfate ions needed for the second cation, Sr^2+, to precipitate is \(3.2\times10^{-6}\,M\).
Key Concepts
Precipitation ReactionSaturation ConcentrationSolubility Constants
Precipitation Reaction
A precipitation reaction occurs when two aqueous solutions are mixed together, leading to the formation of an insoluble solid known as a precipitate. This happens when the product of the ions in solution exceeds the solubility product constant (K_{sp}), which is essentially the threshold concentration beyond which a compound begins to form a solid and precipitate out of solution. In the provided exercise, the precipitation reaction involves the mixing of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) with a solution containing \(\mathrm{Ba}^{2+}\) and \(\mathrm{Sr}^{2+}\). The \(\mathrm{SO}_{4}^{2-}\) ions from \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) interact with the metal cations in the solution to possibly form \(\mathrm{BaSO}_{4}\) or \(\mathrm{SrSO}_{4}\) precipitates.
- Reactants in Solution: These are the substances that dissolve in water to form respective ions. Here, \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) breaks down into \(\mathrm{Na}^{+}\) and \(\mathrm{SO}_{4}^{2-}\) ions.
- Formation of Precipitate: When the \(\mathrm{SO}_{4}^{2-}\) concentration is sufficiently high, either \(\mathrm{BaSO}_{4}\) or \(\mathrm{SrSO}_{4}\) will begin to form as a solid.
- Order of Precipitation: Through the calculation in the exercise, it was shown that \(\mathrm{Ba}^{2+}\) precipitates first as \(\mathrm{BaSO}_{4}\).
Saturation Concentration
Saturation concentration is a key term used to describe the maximum concentration of a solute that can dissolve in a solvent at a given temperature before the solute starts to precipitate. In the exercise given, you were tasked with identifying the saturation concentrations of \(\mathrm{SO}_{4}^{2-}\) in the presence of \(\mathrm{Ba}^{2+}\) and \(\mathrm{Sr}^{2+}\). These values are crucial for determining when each metallic sulfate will start to precipitate from the solution.
- BaSO4 Saturation Concentration: We calculated it to be \(1.1\times10^{-9} \mathrm{M}\).
- SrSO4 Saturation Concentration: Was found to be \(3.2\times10^{-6} \mathrm{M}\).
- Comparing Saturations: A lower saturation concentration indicates a higher tendency to precipitate. Hence, BaSO4 precipitates at a lower concentration of \(\mathrm{SO}_{4}^{2-}\), showing BaSO4 is less soluble in water compared to SrSO4.
Solubility Constants
Solubility constants, represented as \(K_{sp}\), are values that express the equilibrium between the dissolution and precipitation processes of a slightly soluble compound in solution. These constants help in predicting how much of a compound can dissolve before it starts forming a precipitate. In the context of the exercise, we dealt with the solubility constants of two compounds: \(\mathrm{BaSO}_{4}\) with a \(K_{sp}\) of \(1.1 \times 10^{-10}\) and \(\mathrm{SrSO}_{4}\) with a \(K_{sp}\) of \(3.2 \times 10^{-7}\).
- Using \(K_{sp}\): These values allow us to determine the concentrations at which particular ions will start precipitating out whenever their product exceeds the \(K_{sp}\) value.
- Precise Calculations: From the solubility constants, we derive the necessary formulas to calculate when precipitation occurs, and, importantly, the conditions needed for selective precipitation.
- Interpreting \(K_{sp}\) Values: A smaller \(K_{sp}\) indicates a less soluble compound. In our case, BaSO4 is less soluble than SrSO4, evidenced by its lower solubility constant.
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