Problem 54 - Chemistry

Question

A solution contains \(1.0 \times 10^{-6} M \mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) and \(5.0 \times 10^{-7} M\) \(\mathrm{K}_{3} \mathrm{PO}_{4} .\) Will \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) precipitate? \(\left[K_{\mathrm{sp}} \text { for } \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}=1.0 \times 10^{-31} . ] \right.\)

Step-by-Step Solution

Answer

The ionic product, \(Q_{sp} = 1.25 \times 10^{-34}\), is less than the solubility product constant, \(K_{sp} = 1.0 \times 10^{-31}\). Therefore, no precipitation of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) will occur in the given solution.

Step 1: Write the balanced chemical equation for the reaction and expression for \(K_{sp}\)

: First, let's write the balanced chemical equation for the formation of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) solid: \(3 \mathrm{Sr}^{2+}(aq) + 2 \mathrm{PO}_{4}^{3-}(aq) \rightleftharpoons \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s)\) Now we can write the expression for \(K_{sp}\) of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\): \(K_{sp} = [\mathrm{Sr}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2}\) The given value of \(K_{sp}\) for \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) is \(1.0 \times 10^{-31}\).

Step 2: Calculate the initial concentrations of \(\mathrm{Sr}^{2+}\) and \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions

: The given initial concentration for the \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) solution is \(1.0 \times 10^{-6} \mathrm{M}\) and for the \(\mathrm{K}_{3} \mathrm{PO}_{4}\) solution is \(5.0 \times 10^{-7} \mathrm{M}\). Since one mole of \(\mathrm{Sr}\left(\mathrm{NO}_{3}\right)_{2}\) dissociates completely to produce one mole of \(\mathrm{Sr}^{2+}\) ions, the initial concentration of \(\mathrm{Sr}^{2+}\) ions in the solution is also \(1.0 \times 10^{-6} \mathrm{M}\). Likewise, one mole of \(\mathrm{K}_{3} \mathrm{PO}_{4}\) dissociates completely to produce one mole of \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions, so the initial concentration of \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions is \(5.0 \times 10^{-7} \mathrm{M}\).

Step 3: Calculate the ionic product and compare to \(K_{sp}\)

: Now, we can calculate the ionic product of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) using the initial concentrations of \(\mathrm{Sr}^{2+}\) and \(\left(\mathrm{PO}_{4}\right)^{3-}\) ions: \(Q_{sp} = [\mathrm{Sr}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2} = (1.0 \times 10^{-6})^{3}(5.0 \times 10^{-7})^{2} = 1.25 \times 10^{-34}\) Now, we need to compare the ionic product (\(Q_{sp}\)) with the solubility product constant (\(K_{sp}\)): If \(Q_{sp} > K_{sp}\), then precipitation will occur. If \(Q_{sp} < K_{sp}\), then no precipitation will occur. In this case: \(Q_{sp} = 1.25 \times 10^{-34} < K_{sp} = 1.0 \times 10^{-31}\) Since \(Q_{sp} < K_{sp}\), no precipitation of \(\mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}\) will occur.

Key Concepts

Ksp (Solubility Product Constant)

The solubility product constant, denoted as \( K_{sp} \), is a special type of equilibrium constant used for saturated solutions of salts. It indicates the extent to which a solid can dissolve in water, determining the product of the concentrations of the ions that are in equilibrium with the solid salt. In simpler terms, it tells us how much of a solid can dissolve in a solution before reaching saturation, where no more can dissolve, and any excess forms a precipitate.

Understanding \( K_{sp} \) is crucial in predicting whether a precipitation reaction will occur in a given set of conditions. For example, if the \( K_{sp} \) value for a certain salt is very low, it means that the compound is not highly soluble in water. In our case with \( \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2} \), the \( K_{sp} \) is \( 1.0 \times 10^{-31} \), indicating that it is highly insoluble.

The key role of \( K_{sp} \) is to compare it against the ionic product. If the ionic product, \( Q_{sp} \), is greater than \( K_{sp} \), the solution can no longer hold the ions in solution, causing precipitation. If \( Q_{sp} \) is less, no precipitation occurs.

Ionic Product

The ionic product, often represented as \( Q_{sp} \), is calculated similarly to the solubility product constant but uses the actual concentrations of the ions present in the solution, rather than those at equilibrium. It helps determine the formation of a precipitate.

To find \( Q_{sp} \), multiply the concentration of the contributing ions, taking into account their stoichiometric coefficients from the balanced chemical equation. Using our exercise as an example, the ionic product for \( \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2} \) is calculated as \( Q_{sp} = [\mathrm{Sr}^{2+}]^{3}[\mathrm{PO}_{4}^{3-}]^{2} \). Substituting in the initial ion concentrations gives us a \( Q_{sp} \) of \( 1.25 \times 10^{-34} \).

It's important to remember:

• If \( Q_{sp} > K_{sp} \), precipitation occurs, as the solution is supersaturated.
• If \( Q_{sp} < K_{sp} \), the ions remain dissolved, meaning no precipitate will form as the solution is unsaturated.
  
• If \( Q_{sp} = K_{sp} \), the system is exactly at equilibrium, and it represents a saturated solution where a dynamic balance of dissolution and precipitation occurs.

This comparison is pivotal in predicting the outcome of mixing particular ionic solutions in chemistry.

Balanced Chemical Equations

Balanced chemical equations are fundamental to understanding reactions, including precipitation. They ensure that the number of atoms for each element is the same on both sides of the equation, reflecting the conservation of mass. Balancing is especially important in cases of precipitation reactions, as it helps in determining the stoichiometry of the ions involved.

In the problem provided, the balanced equation for the formation of strontium phosphate, \( \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2} \), is:

• \( 3 \mathrm{Sr}^{2+}(aq) + 2 \mathrm{PO}_{4}^{3-}(aq) \rightleftharpoons \mathrm{Sr}_{3}\left(\mathrm{PO}_{4}\right)_{2}(s) \)

This equation shows that three moles of strontium ions react with two moles of phosphate ions to form solid strontium phosphate. Balancing such equations is critical to calculating \( K_{sp} \) and \( Q_{sp} \), as it tells us how many moles of each ion react and end up in the solid state.

By balancing equations, we also ensure that we use the correct ratios when calculating ionic products or reacting quantities in solution chemistry. It simplifies the complexity of chemical reactions by providing a clear framework of reactants converting to products, crucial in understanding the solution's chemistry accurately.