Question

A student prepared a saturated solution of \(\mathrm{CaCrO}_{4}\) and found that when \(156 \mathrm{~mL}\) of the solution was evaporated, \(0.649 \mathrm{~g}\) of \(\mathrm{CaCrO}_{4}\) was left behind. What is the value of \(K_{\mathrm{sp}}\) for this salt?

Short answer

The Ksp value for CaCrO_4 is calculated by multiplying the concentrations of the calcium and chromate ions in solution, which we determine from the mass of CaCrO_4 left behind and the initial volume of the saturated solution.

Step-by-step solution

Understanding the Concept of Ksp

The solubility product constant, or Ksp, is the equilibrium constant for a solid substance dissolving in an aqueous solution. It represents the level at which a solute dissolves in solution. The formula for Ksp of a general salt, AB, is Ksp = [A^+][B^-], where [A^+] is the concentration of the cation and [B^-] is the concentration of the anion.

Deduce Molar Mass of the Salt

Find the molar mass of CaCrO_4 by adding the atomic masses of Calcium (Ca), Chromium (Cr), and four Oxygen (O) atoms. The atomic masses are approximately 40 g/mol for Ca, 52 g/mol for Cr, and 16 g/mol for each O atom.

Calculate the Number of Moles of Salt

Calculate the moles of CaCrO_4 by dividing the mass of the precipitated CaCrO_4 by its molar mass. Use the formula:n = mass / molar mass

Determine the Concentration of Ions in Solution

Since there is a 1:1 molar ratio between CaCrO_4 and its ions in solution (Ca^2+ and CrO_4^2-), the concentration of each ion is equal to the moles of CaCrO_4 divided by the volume of the solution in liters. Use the formula:[ion] = moles / volume (L)

Calculate the Ksp Value

Knowing the concentrations of the ions, calculate the Ksp for CaCrO_4 by multiplying the concentrations of the ions. Use the formula:Ksp = [Ca^2+][CrO_4^2-]

Convert the Volume to Liters

Since the volume of the solution is given in mL, convert it to liters by dividing by 1000.volume (L) = volume (mL) / 1000

Plug Values into the Ksp Expression

Using the concentrations from Step 4 and the molarity calculated in Step 3, plug these values into the expression from Step 5 to find the Ksp value for CaCrO_4.

Key concepts

Equilibrium Constant

Understanding the equilibrium constant is crucial when studying chemical equilibria. It is a number that expresses the ratio of the concentration of the products to the concentration of the reactants for a reversible chemical reaction at equilibrium, each raised to the power of their coefficients in the balanced equation.

For the dissolution of solids in water, this constant is referred to as the solubility product constant, or Ksp. This is a special case of the equilibrium constant that applies to solutes that dissociate into ions in solution. The Ksp is particularly useful because it indicates the extent to which a compound will dissolve in water and provides insights into the solution's saturation level.

When we talk about the Ksp for \(\mathrm{CaCrO}_{4}\), it represents the equilibrium state where \(\mathrm{CaCrO}_{4}\) is in a dynamic balance between the solid and dissolved ion forms \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\). This balance is captured by the expression \(K_{sp} = [\mathrm{Ca}^{2+}][\mathrm{CrO}_{4}^{2-}]\), where the square brackets denote the concentration of ions in molarity.

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry that represents the mass of one mole of a substance, typically expressed in grams per mole (g/mol). To calculate this value, sum the atomic masses of all the atoms in the compound's formula.

In the context of the problem provided, where the salt is \(\mathrm{CaCrO}_{4}\), the molar mass calculation would involve adding the atomic masses of one calcium (Ca), one chromium (Cr), and four oxygen (O) atoms. Since the atomic weights of Ca, Cr, and O are approximately 40 g/mol, 52 g/mol, and 16 g/mol, respectively, the molar mass of \(\mathrm{CaCrO}_{4}\) is the sum of these weights.

The accurate determination of molar mass is essential when converting between mass and moles, which is a cornerstone for understanding the quantitative aspects of chemical reactions, including those involving solubility and Ksp.

Concentration of Ions

Ion concentration is a measure of the amount of dissolved ions present in a particular volume of solution. It is usually expressed as molarity, which is moles of solute per liter of solution (mol/L). For a saturated solution in equilibrium with a precipitate, the ion concentration can be directly related to the solubility product constant (Ksp).

After determining the number of moles of \(\mathrm{CaCrO}_{4}\) that precipitated from the solution, the next step is to establish the concentration of the constituent ions \(\mathrm{Ca}^{2+}\) and \(\mathrm{CrO}_{4}^{2-}\). Since for each mole of \(\mathrm{CaCrO}_{4}\) that dissolves, one mole of \(\mathrm{Ca}^{2+}\) and one mole of \(\mathrm{CrO}_{4}^{2-}\) ions are formed, their concentrations are equal and directly calculated from the moles of \(\mathrm{CaCrO}_{4}\) as derived from the precipitated mass.

Molarity

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute divided by the volume of solution in liters. Represented by the symbol 'M', its unit of measure is moles per liter (mol/L). For example, a 1 M solution contains 1 mole of solute per liter of solution.

In the given exercise, molarity plays a significant role in the calculation of the solubility product constant (Ksp). The solubility of \(\mathrm{CaCrO}_{4}\) is expressed in terms of the molarity of its ions. The concentration of these ions is obtained by dividing the moles of \(\mathrm{CaCrO}_{4}\) by the volume of the solution, which must be in liters. To ensure accuracy, it's crucial to convert the volume from milliliters to liters before doing this calculation.

As the volume of the solution influences the molarity, and consequently, the Ksp value, proper conversion and precise molarity determination are key steps in solving solubility equilibrium problems.