Question

The \(K_{\mathrm{sp}}\) value for radium sulfate, \(\mathrm{RaSO}_{4}\), is \(4.2 \times 10^{-11} .\) If \(25 \mathrm{mg}\) of radium sulfate is placed in \(1.00 \times 10^{2} \mathrm{mL}\) of water, does all of it dissolve? If not, how much dissolves?

Short answer

Not all of the radium sulfate dissolves; only about 0.0946 mg dissolves.

Step-by-step solution

Write the dissolution equation

The dissolution of radium sulfate in water can be represented by the following equation:\[\text{RaSO}_4 (s) \rightleftharpoons \text{Ra}^{2+} (aq) + \text{SO}_4^{2-} (aq)\]

Define the solubility product expression

The solubility product expression for the dissolution is given by:\[K_{\text{sp}} = [\text{Ra}^{2+}][\text{SO}_4^{2-}]\]

Calculate molar mass of radium sulfate

First, we find the molar mass of radium sulfate (\(\text{RaSO}_4\)). Radium (Ra) has a molar mass of approximately 226.0 g/mol, sulfur (S) is about 32.07 g/mol, and oxygen (O) is about 16.00 g/mol for each of the 4 oxygen atoms:\[\text{Molar Mass of RaSO}_4 = 226.0 + 32.07 + 4 \times 16.00 = 146.07\, \text{g/mol}\]

Convert mass to moles

Convert the given mass of radium sulfate to moles using its molar mass:\[\text{Moles of RaSO}_4 = \frac{25 \text{ mg} \times 10^{-3} \text{ g/mg}}{146.07 \text{ g/mol}}\approx 0.171 \times 10^{-3} \text{ mol}\]

Calculate concentration in the solution

Convert the volume of water to liters:\[100 \text{ mL} = 0.1 \text{ L}\]Compute the concentration of radium sulfate:\[\text{Concentration of RaSO}_4 = \frac{0.171 \times 10^{-3}}{0.1} = 0.00171 \text{ mol/L}\]

Use solubility product to find maximum solubility

Let \(s\) be the solubility of \(\text{RaSO}_4\). Then, at equilibrium:\[K_{\text{sp}} = s^2 = 4.2 \times 10^{-11}\]So, \(s = \sqrt{4.2 \times 10^{-11}}\approx 6.48 \times 10^{-6}\, \text{mol/L} \).

Compare solubility to concentration

The calculated solubility of \(\text{RaSO}_4\) is \(6.48 \times 10^{-6} \text{ mol/L}\), which is far less than the concentration provided by dissolving 25 mg of radium sulfate (\(0.00171 \text{ mol/L}\)). Therefore, not all of the radium sulfate will dissolve.

Calculate final dissolved mass

Find the amount of dissolved radium sulfate given its solubility:\[\text{Mass of dissolved RaSO}_4 = s \times 0.1 \text{ L} \times 146.07 \text{ g/mol}\approx 6.48 \times 10^{-6} \times 0.1 \times 146.07\approx 9.46 \times 10^{-5} \text{ g}\]Convert to mg:\[9.46 \times 10^{-5} \text{ g} = 0.0946 \text{ mg}\]

Key concepts

Solubility Product Constant (Ksp)

The Solubility Product Constant, or \( K_{sp} \), is a vital concept in understanding the solubility of sparingly soluble salts in a solution. It represents the maximum concentration of ions that can dissolve in a solution before reaching a state of equilibrium. For radium sulfate, \( \text{RaSO}_4 \), this constant is given as \( 4.2 \times 10^{-11} \).
This value tells us how much of a compound can dissolve in water to produce its constituent ions. The lower the \( K_{sp} \) value, the less soluble the substance. In the case of radium sulfate, the very low \( K_{sp} \) indicates that it dissolves only slightly in water. Understanding \( K_{sp} \) helps in predicting whether a precipitate will form in a solution or how much of a material will dissolve over time.

Dissolution Equation

In any dissolution process, a chemical equation is used to represent the transformation of a solid into its constituent ions in solution. For radium sulfate, \( \text{RaSO}_4 \), this is expressed in a reversible reaction as follows:
\[ \text{RaSO}_4 (s) \rightleftharpoons \text{Ra}^{2+} (aq) + \text{SO}_4^{2-} (aq) \]
This equation illustrates the breakdown of solid radium sulfate into radium and sulfate ions when added to water. Understanding this reaction is crucial as it forms the basis for calculating various parameters like the solubility product and the extent to which the compound dissolves in water. Such dissolution equations make it easier to visualize the process and apply stoichiometric relationships.

Molar Mass Calculation

Calculating the molar mass of a compound is essential for converting between grams and moles, which is a key step in many chemical calculations. For radium sulfate, \( \text{RaSO}_4 \), the molar mass can be calculated by adding the atomic masses of its constituent elements:
- Radium (Ra): approximately 226.0 g/mol
- Sulfur (S): approximately 32.07 g/mol
- Oxygen (O): 16.00 g/mol per atom, and there are 4 oxygen atoms.
Hence, the molar mass is calculated as:
\[ 226.0 + 32.07 + (4 \times 16.00) = 146.07\, \text{g/mol} \]This value is crucial for finding how many moles are present in a given mass of radium sulfate, enabling further calculations like solution concentration and dissolved amounts.

Solution Concentration Calculation

To find out how much of a solute is present in a solution, calculating its concentration is necessary. This is often expressed in terms of moles per liter (mol/L). For the given radium sulfate scenario, we need to compute the concentration of \( \text{RaSO}_4 \) in the water.
First, convert the mass of radium sulfate (25 mg) to grams: 25 mg is 0.025 g. Using the molar mass we've calculated (146.07 g/mol), convert this mass to moles:
\[ \text{Moles of } \text{RaSO}_4 = \frac{0.025 \times 10^{-3}\, \text{g}}{146.07\, \text{g/mol}} \approx 0.171 \times 10^{-3} \text{ mol} \]
Then, convert the solution volume from milliliters to liters: 100 mL is 0.1 L. Now, calculate the concentration:
\[ \text{Concentration} = \frac{0.171 \times 10^{-3}\, \text{mol}}{0.1\, \text{L}} = 0.00171\, \text{mol/L} \]
This tells us the amount of radium sulfate currently in the solution, helping compare it with the solubility limit to determine how much dissolves entirely.