Problem 89 - Chemistry Matter and Change

Question

Compare and Contrast Which of the two solids, calcium phosphate or iron(III) phosphate, has the greater molar solubility? \(K_{\mathrm{sp}}\left(\mathrm{Ca}_{3}\left(\mathrm{PO}_{4}\right)_{2}\right)=1.2 \times 10^{-29}; \) \(K_{\mathrm{sp}}\left(\mathrm{FePO}_{4}\right)=1.0 \times 10^{-22} .\) Which compound has the greater solubility, expressed in grams per liter?

Step-by-Step Solution

Answer

Calcium phosphate is more soluble than iron(III) phosphate.

Step 1: Understand the definition of molar solubility

Molar solubility refers to the number of moles of a solute that can dissolve in 1 liter of solution before the solution becomes saturated.

Step 2: Write the dissolution equation for each compound

For calcium phosphate (\( \text{Ca}_3(\text{PO}_4)_2 \)), the dissolution equation is:\[ \text{Ca}_3(\text{PO}_4)_2 (s) \rightleftharpoons 3\text{Ca}^{2+} (aq) + 2\text{PO}_4^{3-} (aq) \]For iron(III) phosphate (\( \text{FePO}_4 \)), the dissolution equation is:\[ \text{FePO}_4 (s) \rightleftharpoons \text{Fe}^{3+} (aq) + \text{PO}_4^{3-} (aq) \]

Step 3: Express the solubility products in terms of molar solubility

For calcium phosphate, if \( s \) is the molar solubility, then:\[ K_{sp} = [\text{Ca}^{2+}]^3 [\text{PO}_4^{3-}]^2 = (3s)^3 (2s)^2 \]For iron(III) phosphate, with \( s \) as the molar solubility, we have:\[ K_{sp} = [\text{Fe}^{3+}][\text{PO}_4^{3-}] = s^2 \]

Step 4: Solve for molar solubility

For calcium phosphate, using \( K_{sp} = 1.2 \times 10^{-29} \):\[ (3s)^3 (2s)^2 = 1.2 \times 10^{-29} \]\[ 108s^5 = 1.2 \times 10^{-29} \]\[ s = \left(\frac{1.2 \times 10^{-29}}{108}\right)^{1/5} \]For iron(III) phosphate, using \( K_{sp} = 1.0 \times 10^{-22} \):\[ s^2 = 1.0 \times 10^{-22} \]\[ s = \sqrt{1.0 \times 10^{-22}} \]

Step 5: Calculate and compare the solubilities

For calcium phosphate: \[ s = \left(\frac{1.2 \times 10^{-29}}{108}\right)^{1/5} \approx 2.0 \times 10^{-6} \text{ mol/L} \]For iron(III) phosphate:\[ s = \sqrt{1.0 \times 10^{-22}} = 1.0 \times 10^{-11} \text{ mol/L} \]Thus, calcium phosphate has a greater molar solubility than iron(III) phosphate.

Step 6: Convert molar solubility to grams per liter

For calcium phosphate, calculate the mass per liter:\( s = 2.0 \times 10^{-6} \text{ mol/L} \), Molar Mass of \( \text{Ca}_3(\text{PO}_4)_2 \approx 310.18 \text{ g/mol} \)\[ \Rightarrow 2.0 \times 10^{-6} \text{ mol/L} \times 310.18 \text{ g/mol} = 6.2 \times 10^{-4} \text{ g/L} \]For iron(III) phosphate, calculate the mass per liter:\( s = 1.0 \times 10^{-11} \text{ mol/L} \), Molar Mass of \( \text{FePO}_4 \approx 150.82 \text{ g/mol} \)\[ \Rightarrow 1.0 \times 10^{-11} \text{ mol/L} \times 150.82 \text{ g/mol} = 1.5 \times 10^{-9} \text{ g/L} \]

Step 7: Compare solubilities in grams per liter

Calcium phosphate: \( 6.2 \times 10^{-4} \text{ g/L} \)Iron(III) phosphate: \( 1.5 \times 10^{-9} \text{ g/L} \)Calcium phosphate has a greater solubility than iron(III) phosphate when expressed in grams per liter.

Key Concepts

Molar Solubility

Molar solubility is the measure of a solute that can dissolve in a particular amount of solvent to form a saturated solution, usually expressed in moles per liter (mol/L). Think of it as the point at which a solution can no longer dissolve more of the solute without some of it staying in a solid state. Molar solubility is crucial in determining how much of a compound can be dissolved in water or any other solvent.

In this context, comparing calcium phosphate and iron(III) phosphate, molar solubility provides insight into which of the two can dissolve more efficiently. A higher molar solubility indicates that a greater quantity of the compound can dissolve in the solution. For example, in our exercise, calculating the molar solubility helps us determine that calcium phosphate has a higher capacity for dissolution compared to iron(III) phosphate. This is important for applications that require predictable solute concentrations in solution.

Solubility Product (Ksp)

The solubility product constant, represented as \( K_{sp} \), is a numerical value that reflects the solubility of a compound in terms of its dissociated ions. It's an equilibrium constant for the dissolution of a sparingly soluble salt. A higher \( K_{sp} \) value indicates greater solubility.

For each dissolution reaction, such as calcium phosphate or iron(III) phosphate in water, the \( K_{sp} \) equation is set up based on the concentrations of the ions produced. The general formula takes the arrangement of the dissolution equation into account. For calcium phosphate, we have:

• \( ext{Ca}_3( ext{PO}_4)_2 (s) \rightarrow 3 ext{Ca}^{2+} + 2 ext{PO}_4^{3-} \)

Thus, \( K_{sp} = [ ext{Ca}^{2+}]^3 [ ext{PO}_4^{3-}]^2 \).

In simpler terms, \( K_{sp} \) allows chemists to predict the concentrations of different ions in a saturation equilibrium, helping compare the solubility levels of different compounds more accurately.

Calcium Phosphate

Calcium phosphate, with the formula \( ext{Ca}_3( ext{PO}_4)_2 \), is an inorganic compound used broadly in fertilizers and certain types of glasses. It has a relatively low solubility in water, making it ideal for applications requiring slow release of calcium and phosphate ions.

In the exercise, we calculated the molar solubility of calcium phosphate to be \( 2.0 \times 10^{-6} \text{ mol/L} \), which is relatively higher compared to iron(III) phosphate. This means that, when dissolved in water, calcium phosphate can provide more calcium and phosphate ions than iron(III) phosphate can provide its ions.

When expressed in grams per liter, the solubility of calcium phosphate equates to \( 6.2 \times 10^{-4} \text{ g/L} \). The molar mass of calcium phosphate is essential here, as it helps convert the molar solubility (mol/L) into a more tangible quantity (g/L), making it easier to compare with other compounds.

Iron(III) Phosphate

Iron(III) phosphate, with the chemical formula \( ext{FePO}_4 \), is often used in agriculture as a slug and snail bait due to its low solubility. The dissolution equation outlines how it dissociates into iron and phosphate ions in water:

• \( ext{FePO}_4 (s) \rightarrow ext{Fe}^{3+} + ext{PO}_4^{3-} \)

The exercise and its calculations identify the solubility product \( K_{sp} \) as relatively higher than that of calcium phosphate, yet its molar solubility is significantly lower.

The molar solubility calculated for iron(III) phosphate is \( 1.0 \times 10^{-11} \text{ mol/L} \), denoting much lower solubility compared to calcium phosphate. Expressing this in grams per liter, we find it's only \( 1.5 \times 10^{-9} \text{ g/L} \). This low solubility correlates with its practical uses, where slow dissolution is often desirable.