Question

The \(K_{\mathrm{sp}}\) of \(\mathrm{SrSO}_{4}\) is \(7.6 \times 10^{-7}\). Should precipitation occur when \(25.0 \mathrm{~mL}\) of \(1.0 \times 10^{-3} \mathrm{M} \mathrm{SrCl}_{2}\) solution are mixed with \(15.0 \mathrm{~mL}\) of \(2.0 \times 10^{-3} M \mathrm{Na}_{2} \mathrm{SO}_{4}\) ? Show proof.

Short answer

Calculate ion concentrations after mixing and compare Q with K_{\text{sp}} to determine if precipitation occurs.

Step-by-step solution

- Calculate the Final Concentrations

First, determine the final concentrations of \text{Sr}^{2+} and \text{SO}_4^{2-} ions after mixing the solutions. The total volume after mixing the solutions is \[ V_{\text{total}} = 25.0 \text{ mL} + 15.0 \text{ mL} = 40.0 \text{ mL} \] The concentration of \text{Sr}^{2+} after mixing is given by the dilution formula: \[ [\text{Sr}^{2+}] = \frac{(25.0 \text{ mL}) \times (1.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} \] Calculate this to find the concentration of \text{Sr}^{2+}. Similarly, calculate the concentration of \text{SO}_4^{2-} after mixing: \[ [\text{SO}_4^{2-}] = \frac{(15.0 \text{ mL}) \times (2.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} \]

- Calculate the Ion Product (Q)

Next, calculate the ion product (\text{Q}) using the concentrations obtained in Step 1:\[ Q = [\text{Sr}^{2+}] \times [\text{SO}_4^{2-}] \] Substitute the values obtained from the previous step to find Q.

- Compare Ion Product (Q) and Solubility Product (K_sp)

Compare the ion product (Q) with the given solubility product (K_{\text{sp}}): \[ K_{\text{sp}} = 7.6 \times 10^{-7} \] Determine if precipitation occurs by comparing Q and K_{\text{sp}}. If Q > K_{\text{sp}}, precipitation will occur. If Q ≤ K_{\text{sp}}, no precipitation will occur.

Key concepts

Solubility Product Constant

The solubility product constant \(K_{sp}\) quantifies the solubility of a sparingly soluble ionic compound in water. It is a type of equilibrium constant specific to the solubility of salts.
When a salt dissolves in water, it dissociates into its constituent ions. For strontium sulfate \((SrSO_4)\), the dissolution can be represented as follows:

\[ SrSO_4 \rightleftharpoons Sr^{2+} + SO_4^{2-} \]
The solubility product constant \(K_{sp}\) is then given by the product of the molar concentrations of these ions at equilibrium:
\[ K_{sp} = [Sr^{2+}][SO_4^{2-}] \]
For \(SrSO_4\), \(K_{sp} = 7.6 \times 10^{-7}\).

Understanding \(K_{sp}\) is crucial as it enables us to predict whether a precipitate will form when two solutions are mixed. By comparing \(K_{sp}\) with the ion product, we can decide if a precipitation reaction occurs.

Ion Product

The ion product \(Q\) is very similar to the solubility product constant \(K_{sp}\), but it applies to the current concentrations of the ions in the solution, rather than the equilibrium concentrations.
To calculate \(Q\), multiply the molar concentrations of the ions present in the solution at any given time:
\[ Q = [Sr^{2+}][SO_4^{2-}] \]
The values for \([Sr^{2+}]\) and \([SO_4^{2-}]\) are determined from the initial concentrations and volumes of the mixed solutions.
In our exercise:
\[ [Sr^{2+}] = \frac{(25.0 \text{ mL}) \times (1.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} = 6.25 \times 10^{-4} \text{ M} \]
\[ [SO_4^{2-}] = \frac{(15.0 \text{ mL}) \times (2.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} = 7.5 \times 10^{-4} \text{ M} \]
Substitute these values into the formula for \(Q\) to find its value.

Precipitation Reaction

A precipitation reaction occurs when the ion product \(Q\) exceeds the solubility product constant \(K_{sp}\).
If \(Q > K_{sp}\), the solution is supersaturated, and the excess ions will combine to form a precipitate and restore equilibrium.
In this exercise, the calculated \(Q\) from the ion concentrations:
\[ Q = (6.25 \times 10^{-4} \text{ M}) \times (7.5 \times 10^{-4} \text{ M}) = 4.69 \times 10^{-7} \]
If we compare \(Q\) to \(K_{sp}\):
\[Q = 4.69 \times 10^{-7} < K_{sp} = 7.6 \times 10^{-7} \]
Because \(Q < K_{sp}\), no precipitation occurs. The ions remain dissolved in the solution.
Understanding this relationship helps in predicting the solubility behavior of various salts under different conditions.

Concentration Calculation

Accurate concentration calculation is integral for determining the ion product \(Q\) and eventually assessing precipitation.
In dilutions, concentration changes when solutions are mixed. We use the formula:

\[ [\text{final}] = \frac{(\text{initial volume}) \times (\text{initial concentration})}{\text{total volume}} \]

For strontium ions \(Sr^{2+}\):
\[ [Sr^{2+}] = \frac{(25.0 \text{ mL}) \times (1.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} = 6.25 \times 10^{-4} \text{ M} \]
For sulfate ions \(SO_4^{2-}\):
\[ [SO_4^{2-}] = \frac{(15.0 \text{ mL}) \times (2.0 \times 10^{-3} \text{ M})}{40.0 \text{ mL}} = 7.5 \times 10^{-4} \text{ M} \]
Using these concentrations aids the calculation of \(Q\) and prediction of precipitation, which are key aspects of chemical equilibria.
Proper concentration calculation ensures accurate predictions in chemical reactions and solubility scenarios.