Question

A solution in equilibrium with solid barium phosphate is found to have a barium ion concentration of \(5.0 \times 10^{-4} \mathrm{M}\) and a \(K_{\mathrm{sp}}\) of \(3.4 \times 10^{-23} .\) Calculate the concentration of phosphate ion.

Short answer

[\text{PO}_4^{3-}] \approx 1.65 \times 10^{-7} \text{ M}

Step-by-step solution

- Write down the dissociation equation

The dissociation of barium phosphate in water can be represented by the following chemical equation:\[ \text{Ba}_3(\text{PO}_4)_2 (s) \leftrightharpoons 3\text{Ba}^{2+} (aq) + 2\text{PO}_4^{3-} (aq) \]

- Write the expression for the solubility product

The solubility product constant (\(K_{sp}\)) expression for barium phosphate is given by:\[ K_{sp} = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2 \]

- Substitute the given values into the expression

We are given \( [\text{Ba}^{2+}] = 5.0 \times 10^{-4} \text{ M} \) and \(K_{sp} = 3.4 \times 10^{-23}\). Substituting these values, we get:\[ 3.4 \times 10^{-23} = (5.0 \times 10^{-4})^3 [\text{PO}_4^{3-}]^2 \]

- Solve for the concentration of phosphate ion

First, calculate \((5.0 \times 10^{-4})^3\):\[ (5.0 \times 10^{-4})^3 = 1.25 \times 10^{-10} \]Next, substitute this value back into the expression:\[ 3.4 \times 10^{-23} = 1.25 \times 10^{-10} [\text{PO}_4^{3-}]^2 \]Solve for \([\text{PO}_4^{3-}]^2\):\[ [\text{PO}_4^{3-}]^2 = \frac{3.4 \times 10^{-23}}{1.25 \times 10^{-10}} \]Calculate the right-hand side:\[ [\text{PO}_4^{3-}]^2 \approx 2.72 \times 10^{-13} \]Finally, take the square root of both sides to solve for \([\text{PO}_4^{3-}]\):\[ [\text{PO}_4^{3-}] = \sqrt{2.72 \times 10^{-13}} \approx 1.65 \times 10^{-7} \text{ M} \]

Key concepts

Chemical Equilibrium

Chemical equilibrium is a state in a chemical reaction where the rates of the forward and reverse reactions are equal. At this point, the concentrations of reactants and products remain constant over time. This doesn't mean the reactions stop; instead, they continue to occur at the same rate, creating a stable balance. Understanding equilibrium is important in solubility problems because it explains how a solution can be saturated with a solute, meaning that no more solute can dissolve at a given temperature without the system being disturbed.

Dissociation Equation

A dissociation equation shows how a compound breaks down into its component ions in a solution. For example, barium phosphate (\text{Ba}_3(\text{PO}_4)_2) dissociates in water as follows:

\(\text{Ba}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3\text{Ba}^{2+} (aq) + 2\text{PO}_4^{3-} (aq) \)
This equation tells us that one formula unit of barium phosphate produces three barium ions (\text{Ba}^{2+}) and two phosphate ions (\text{PO}_4^{3-}) in solution. Knowing how a compound dissociates helps to set up equations for calculating equilibrium concentrations, crucial for solving solubility product constant problems.

Solute Concentration

Solute concentration refers to how much solute is present in a solution. It is often expressed in molarity (M), which is moles of solute per liter of solution. In our exercise, the concentration of barium ion is given as \(5.0 \times 10^{-4} \text{ M}\). This is a key piece of information required for solving equilibrium problems because concentrations of ions are used to calculate the solubility product. Understanding how to read and use these concentrations is essential for making accurate calculations.

Solubility Product Constant Calculation

The solubility product constant (\text{K}_\text{sp}) quantifies the solubility of a compound under specific conditions. For barium phosphate, the expression is:
\[ K_{sp} = [\text{Ba}^{2+}]^3 [\text{PO}_4^{3-}]^2 \]
To solve for the phosphate ion concentration, follow these steps:
1. Substitute the given barium ion concentration and the \text{K}_\text{sp} value into the equation:
\[ 3.4 \times 10^{-23} = (5.0 \times 10^{-4})^3 [\text{PO}_4^{3-}]^2 \]
2. Calculate \( (5.0 \times 10^{-4})^3 \), which equals \( 1.25 \times 10^{-10} \).
3. Substitute and solve for \( [\text{PO}_4^{3-}]^2 \):
\[ [\text{PO}_4^{3-}]^2 = \frac{3.4 \times 10^{-23}}{1.25 \times 10^{-10}} \ [\text{PO}_4^{3-}]^2 \thickapprox 2.72 \times 10^{-13} \]
4. Take the square root to find \ [\text{PO}_4^{3-}]:
\[ [\text{PO}_4^{3-}] \thickapprox 1.65 \times 10^{-7} \text{ M} \].
This shows step-by-step how to find the concentration of an ion in a saturated solution, cementing the relationship between solubility and the solubility product constant.