Question

The \(K_{\mathrm{s} p}\) for \(Q\), a slightly soluble ionic compound composed of \(\mathrm{M}_{2}^{2+}\) and \(\mathrm{X}^{-}\) ions, is \(4.5 \times 10^{-29} .\) The electron configuration of \(\mathrm{M}^{+}\) is \([\mathrm{Xe}] 6 s^{1} 4 f^{14} 5 d^{10}\). The \(\mathrm{X}^{-}\) anion has 54 electrons. What is the molar solubility of \(Q\) in a solution of \(\mathrm{NaX}\) prepared by dissolving \(1.98 \mathrm{~g}\) NaX in \(150 . \mathrm{mL}\) solution?

Short answer

The molar solubility of Q in the NaX solution is \(5.83 \times 10^{-28}\) M.

Step-by-step solution

Determine the initial concentration of X- from NaX

First, we need to calculate the molar mass of NaX and determine the initial concentration of X- ions in the solution. Na = 22.99 g/mol (sodium) X- has 54 electrons with electron configuration [Xe] 6s1 4f14 5d10, which corresponds to the Iodine atom (I). So, I = 126.90 g/mol (iodine) Molar mass of NaX = Na + I = 22.99 g/mol + 126.90 g/mol = 149.89 g/mol Now, calculate the moles of NaX dissolved and convert it into the concentration of X- ions in the 150 mL solution: 1.98 g NaX × (1 mol NaX / 149.89 g) = 0.0132 mol NaX Initial concentration of X- ions: (0.0132 mol) / (0.150 L) = 0.088 M

Set up the Ksp expression and the concentration changes table.

Write out the Ksp expression for Q, considering the dissolution process: \(Q \rightleftharpoons M_{2}^{2+} + 2X^{-}\) Ksp = [M2+][X-]^2 = 4.5 x 10^(-29) Now, set up a table representing the concentrations of M2+ and X- ions before and after the dissolution of Q. | | M2+ | X- | |-----------|---------|----------| | Initial | 0 | 0.088 M | | Change | +s | +2s | | Equilibrium | s | 0.088+2s |

Solve for s.

Plug the concentrations from the table into the Ksp expression and solve for s (molar solubility of Q). \(4.5 \times 10^{-29} = s(0.088+2s)^2\) Since Ksp is very small, we can make the assumption that the change in the concentration of X- ions due to Q's dissolution (2s) is negligible compared to the initial concentration of X- ions (0.088 M). So, we can approximate the expression as: \(4.5 \times 10^{-29} = s(0.088)^2\) Now, solve for s: s = \( \frac{4.5 \times 10^{-29}}{(0.088)^2} \) s = 5.83 × 10^(-28) M

Express the molar solubility of Q

The molar solubility of Q in this solution of NaX is s = 5.83 × 10^(-28) M.

Key concepts

Solubility Product Constant (Ksp)

The concept of the solubility product constant, or Ksp, is pivotal when studying the solubility of sparingly soluble ionic compounds. Ksp is a special type of equilibrium constant that quantifies the degree to which a compound can dissolve in water. It represents the product of the concentrations of the ions in a saturated solution, each raised to the power of their stoichiometric coefficients. The expression for Ksp does not include the concentration of the solid since it is a constant.

Understanding Ksp can help predict whether a precipitate will form in a solution. When the product of the concentrations of the ions in a solution exceeds the Ksp, a precipitate will begin to form until the ion product equals the Ksp value. Meanwhile, if the ion product is less than Ksp, the solution is not saturated and more of the compound can dissolve. Ksp is temperature-dependent and varies for different substances, which is why one needs Ksp values from tables or experiments for specific calculations.

Ionic Compounds

Ionic compounds consist of positive and negative ions held together by the strong electrostatic forces known as ionic bonds. These compounds typically have high melting and boiling points and, when dissolved in water, they dissociate into their constituent ions. The solubility of ionic compounds is influenced by the lattice energy of the solid and the hydration energy of the ions. Some ionic compounds are highly soluble in water, while others are considered slightly soluble or insoluble.

For instance, ionic compounds with common ions can affect each other's solubility through a principle known as the common-ion effect. In the provided problem, the compound Q releases ions M₂²⁺ and X^-. If the solution already contains a high concentration of X^- provided by a soluble salt like NaX, the solubility of Q will be further reduced due to the common-ion effect.

Concentration Calculation

Concentration calculations are used to determine the amount of solute in a given volume of solution and are essential in understanding chemical reactions happening in solutions. Concentration is usually expressed in molarity (M), which is the number of moles of solute per liter of solution. To calculate it, one must first find the number of moles of the solute, which can be done by dividing the mass of the solute by its molar mass. Then, divide the moles of solute by the volume of the solution in liters.

In the example of our ionic compound Q dissolving in a solution of NaX, the initial concentration of X^- is critical. It ensures that the calculation of Q's solubility is precise by considering the common-ion effect. By setting up an ice table and using the Ksp expression, we can subsequently find the molar solubility of Q even in the presence of a common ion. For accuracy, it is important to make justified approximations, such as neglecting the small increases in ion concentration due to the low solubility of Q, which simplifies the mathematics without significantly affecting the result.