Problem 80 - Chemistry The Molecular Science

Question

A solution contains \(0.020 \mathrm{M}\) sodium phosphate and \(0.050 \mathrm{M}\) potassium sulfate. Thus, this solution contains two anion species-phosphate and sulfate. Barium chloride is added to the solution. (a) Two precipitates are possible. Determine which forms first. (b) Calculate the concentration of the first anion species when the second anion species starts to precipitate.

Step-by-Step Solution

Answer

(a) BaSO₄ forms first. (b) [PO₄³⁻] is approximately 9.7 × 10⁻⁴ M.

Step 1: Write the Precipitation Reactions

The first step is to identify the chemical reactions of barium chloride with each of the anions. 1. Reaction with phosphate: \[ 2 \text{PO}_4^{3-} + 3 \text{Ba}^{2+} \rightarrow \text{Ba}_3(\text{PO}_4)_2(s) \] 2. Reaction with sulfate: \[ \text{SO}_4^{2-} + \text{Ba}^{2+} \rightarrow \text{BaSO}_4(s) \]

Step 2: Determine Solubility Products (Ksp)

Next, find or assume the solubility product constants (\(K_{sp}\)) for both precipitates:- \(K_{sp}(\text{Ba}_3(\text{PO}_4)_2) = 1 \times 10^{-13}\)- \(K_{sp}(\text{BaSO}_4) = 1.1 \times 10^{-10}\)These values are necessary to determine when each precipitate starts forming.

Step 3: Calculate Ion Concentrations for Precipitation

Determine the necessary concentration of \(\text{Ba}^{2+}\) ions to start forming each precipitate using the \(K_{sp}\) values:- For phosphate: \[ K_{sp}(\text{Ba}_3(\text{PO}_4)_2) = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2 \] Let \([\text{PO}_4^{3-}] = 0.020 \text{ M}\). Then, \[ 1 \times 10^{-13} = [\text{Ba}^{2+}]^3 (0.020)^2 \] \[ [\text{Ba}^{2+}] = (\frac{1 \times 10^{-13}}{4 \times 10^{-4}})^{1/3} \approx 6.3 \times 10^{-4} \text{ M}\]- For sulfate: \[ K_{sp}(\text{BaSO}_4) = [\text{Ba}^{2+}][\text{SO}_4^{2-}] \] Let \([\text{SO}_4^{2-}] = 0.050 \text{ M}\). Then, \[ 1.1 \times 10^{-10} = [\text{Ba}^{2+}] (0.050) \] \[ [\text{Ba}^{2+}] = \frac{1.1 \times 10^{-10}}{0.050} \approx 2.2 \times 10^{-9} \text{ M}\]

Step 4: Determine Which Precipitate Forms First

Compare the \([\text{Ba}^{2+}]\) ion concentrations required for precipitation. - \([\text{Ba}^{2+}] = 6.3 \times 10^{-4} \text{ M}\) for \(\text{Ba}_3(\text{PO}_4)_2\)- \([\text{Ba}^{2+}] = 2.2 \times 10^{-9} \text{ M}\) for \(\text{BaSO}_4\)Since the concentration for sulfate is much lower, \(\text{BaSO}_4\) will precipitate first.

Step 5: Calculate Concentration of Phosphate Anion When Sulfate Anion Precipitates

Since \(\text{BaSO}_4\) forms first and we know the \([\text{Ba}^{2+}]\) concentration when this happens, we can use it to find \([\text{PO}_4^{3-}]\) at this point:From \[ K_{sp}(\text{Ba}_3(\text{PO}_4)_2) = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2 \], substitute \([\text{Ba}^{2+}] = 2.2 \times 10^{-9} \text{ M}\).\[ 1 \times 10^{-13} = (2.2 \times 10^{-9})^3 [\text{PO}_4^{3-}]^2 \]\[ [\text{PO}_4^{3-}]^2 = \frac{1 \times 10^{-13}}{(2.2 \times 10^{-9})^3} \]\[ [\text{PO}_4^{3-}] = \sqrt{\frac{1 \times 10^{-13}}{1.06 \times 10^{-26}}} \approx 9.7 \times 10^{-4} \text{ M} \]

Key Concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, abbreviated as \(K_{sp}\), is a vital concept when studying precipitation reactions. It's a special equilibrium constant used for sparingly soluble ionic compounds, and it indicates the level at which a compound can dissolve in a solution. Remember that every ionic compound has its own \(K_{sp}\) value, which varies based on the temperature.
In the context of our exercise, we look at two different compounds: barium phosphate \(\text{Ba}_3(\text{PO}_4)_2\) and barium sulfate \(\text{BaSO}_4\). Each has a unique \(K_{sp}\) value. Specifically, \(\text{Ba}_3(\text{PO}_4)_2\) has a \(K_{sp}\) of \(1 \times 10^{-13}\), and \(\text{BaSO}_4\) has a \(K_{sp}\) of \(1.1 \times 10^{-10}\). These values tell us how readily each compound will dissolve in water before starting to precipitate.
When \(\text{Ba}^{2+}\) ions are introduced into the solution, they interact with phosphate and sulfate ions to potentially form solid precipitates. For each precipitate to form, the product of the ion concentrations must exceed the \(K_{sp}\) of that compound. Hence, \(K_{sp}\) is like a threshold; surpass it, and precipitation starts.

Ion Concentration Calculation

Ion concentration calculations are crucial for determining when precipitation will occur in a reaction. Using the solubility product constant together with known ion concentrations, we can deduce the necessary concentration of barium ions \([\text{Ba}^{2+}]\) for each precipitate to start forming.
Let's break it down for our problem:

• For barium phosphate \(\text{Ba}_3(\text{PO}_4)_2\): The expression based on the balanced reaction equation is \(K_{sp} = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2\). With \([\text{PO}_4^{3-}] = 0.020 \, \text{M}\), we can solve for \([\text{Ba}^{2+}]\) to get approximately \(6.3 \times 10^{-4} \, \text{M}\).


• For barium sulfate \(\text{BaSO}_4\): The expression is simpler: \(K_{sp} = [\text{Ba}^{2+}][\text{SO}_4^{2-}]\). Using \([\text{SO}_4^{2-}] = 0.050 \, \text{M}\), the concentration of \([\text{Ba}^{2+}]\) required is \(2.2 \times 10^{-9} \, \text{M}\).

By contrasting these values, we identify which reaction occurs first, as the one requiring the lowest \([\text{Ba}^{2+}]\) concentration will precipitate first.

Barium Chloride Reactions

Barium chloride is a versatile reactant and crucial in our problem as it provides the \(\text{Ba}^{2+}\) ions for the precipitation reactions. Its interactions with other compounds lead to the formation of solid precipitates.
In this exercise, barium chloride reacts with two anions present in the solution: phosphate \((\text{PO}_4^{3-})\) and sulfate \((\text{SO}_4^{2-})\). Here's what happens in each scenario:

• Reaction with phosphate ions leads to the formation of barium phosphate \(\text{Ba}_3(\text{PO}_4)_2\), which has a lower \(K_{sp}\) value compared to barium sulfate, indicating it is less soluble.


• Reaction with sulfate ions creates barium sulfate \(\text{BaSO}_4\), which has a higher \(K_{sp}\), suggesting it will precipitate sooner under the conditions in the solution.

Since barium sulfate requires a lower \([\text{Ba}^{2+}]\) to precipitate than barium phosphate, it forms first. Understanding barium chloride's role in these reactions helps us predict precipitation events effectively and based on the logic that less soluble products will form if a system's saturation is reached.