Question

A handbook lists the solubility of \(\mathrm{CaHPO}_{4}\) as \(0.32 \mathrm{g}\) \(\mathrm{CaHPO}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O} / \mathrm{L}\) and lists \(K_{\mathrm{sp}}\) as \(1 \times 10^{-7}\). (a) Are these data consistent? (That is, are the molar solubilities the same when derived in two different ways?) (b) If there is a discrepancy, how do you account for it?

Short answer

The given solubility data and solubility product constant (Ksp) are not consistent because the calculated Ksp of \(3.4 \times 10^{-6}\) based on the given solubility differs from the provided Ksp of \(1 \times 10^{-7}\). The discrepancy could be due to various factors such as different experimental conditions, rounding errors, or approximations in the calculation.

Step-by-step solution

Convert Solubility into Molarity

First, convert the given solubility from grams per liter into molarity. Molarity is defined as the number of moles of solute per liter of solvent. The molar mass of CaHPO4•2H2O is approximately \(174.2 \, \text{g/mol}\). Thus, the molar solubility is \( \frac{0.32\,g}{174.2\,g/mol}=1.84\times 10^{-3} \,mol/L.\)

Setup Ksp Equation

The equilibrium reaction for \( CaHPO_4 \) dissolving in water is: \(CaHPO_4 \leftrightarrow Ca^{2+} + HPO_4^{-2}\). Therefore, the expression for the solubility product constant would be \(K_{sp}=[Ca^{2+}][HPO_4^{-2}]\). Since the molarities of \(Ca^{2+}\) and \(HPO_4^{-2}\) are equal, we can simplify the equation to \(K_{sp} = (s)^2\), where \(s\) is the molar solubility.

Calculate and Compare with the Given Ksp

Plugging the calculated molar solubility \(s = 1.84\times 10^{-3}\) into the Ksp equation from step 2, we get \(K_{sp} = (1.84\times 10^{-3} \, mol/L )^2 = 3.4\times 10^{-6}\). This value differs from the provided \(K_{sp} = 1 \times 10^{-7}\), which indicates that the given data are not consistent.

Explain the Discrepancy

The discrepancy could result from the original data handbook's different experimental or environmental conditions, rounding errors, or the use of approximations in the calculation. For a detailed explanation, one should refer to the context or specific footnotes of the original data source.

Key concepts

Molar Solubility

In chemistry, molar solubility is how we measure the concentration of a substance when it is dissolved in a liquid. It tells us the number of moles of a compound that can dissolve in a liter of solvent at a specific temperature. This is crucial for understanding how substances interact in solutions.
To calculate molar solubility, you convert the mass of the solute given in grams to moles, using its molar mass. Then, you divide the moles by the volume in liters.
In our example, the solubility of \(\mathrm{CaHPO}_4 \cdot 2 \mathrm{H}_2\mathrm{O}\) is given as 0.32 grams per liter. By converting this using the molar mass of 174.2 g/mol, the molar solubility is found to be \(1.84 \times 10^{-3}\) mol/L. This conversion is a foundational step in comparing real experimental solubility to theoretical calculations using solubility constants.

Solubility Product Constant (Ksp)

The solubility product constant, or \(K_{sp}\), is a measure of how much a compound can dissolve in water to form a saturated solution. It's calculated by multiplying the concentrations of the ions in the solution, each raised to the power of their respective coefficients in the balanced equation.
The equation for the dissolution of \(\mathrm{CaHPO}_4\) is: \(\mathrm{CaHPO}_4 \leftrightarrow \mathrm{Ca}^{2+} + \mathrm{HPO}_4^{-2}\). In this reaction, each mole of \(\mathrm{CaHPO}_4\) produces one mole of \(\mathrm{Ca}^{2+}\) and one mole of \(\mathrm{HPO}_4^{-2}\).
Therefore, the expression for \(K_{sp}\) simplifies to \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{HPO}_4^{-2}] = s^2\), where \(s\) represents the molar solubility. In the example, using the calculated molar solubility \(s = 1.84 \times 10^{-3}\), the theoretical \(K_{sp}\) is \(3.4 \times 10^{-6}\). This value helps us explore if the experimental data matches theoretical expectations.

Inconsistency Explanation in Data

Upon calculating both the theoretical and given values of \(K_{sp}\), a notable inconsistency appears. The calculated \(K_{sp}\) from the molar solubility does not match the handbook's listed \(K_{sp}\) of \(1 \times 10^{-7}\). This discrepancy can arise from several factors.

• Environmental factors such as temperature or pH might vary between experiments, affecting solubility.
• Rounding off can introduce deviations in calculation, as small errors multiply during conversions.
• The handbook data may involve approximations or assumptions that differ from real conditions.

In practice, checking the context or specific conditions under which the handbook data was obtained can shed light on such differences. It's a reminder that theoretical calculations and experimental data must often be understood within the nuances of the experimental setup.