Problem 55 - Chemical Principles

Question

The molarity of \(\mathrm{CrO}_{4}^{2-}\) in a saturated \(\mathrm{Tl}_{2} \mathrm{CrO}_{4}\) solution is \(6.3 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1}\). What is the \(K_{\mathrm{sp}}\) of \(\mathrm{Tl}_{2} \mathrm{CrO}_{4}\) ?

Step-by-Step Solution

Answer

The \(K_{\mathrm{sp}}\) of \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\) is approximately \(1.00 \times 10^{-12}\).

Step 1: Write the Chemical Equation

Write the dissolution equation of the salt in water. Here \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\) dissociates into thallium ions \(\mathrm{Tl}^{+}\) and chromate ions \(\mathrm{CrO}_{4}^{2-}\): \[\mathrm{Tl}_{2}\mathrm{CrO}_{4}(s) \rightleftharpoons 2\mathrm{Tl}^{+}(aq) + \mathrm{CrO}_{4}^{2-}(aq)\]

Step 2: Express the Solubility Product Constant \(K_{\mathrm{sp}}\)

Write the expression for the solubility product constant \(K_{\mathrm{sp}}\) based on the dissociation equation. For the reaction above, the expression is: \[K_{\mathrm{sp}} = [\mathrm{Tl}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\]

Step 3: Determine the Concentration of \(\mathrm{Tl}^{+}\) Ions

Since every mole of \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\) produces two moles of \(\mathrm{Tl}^{+}\), the molarity of \(\mathrm{Tl}^{+}\) ions will be twice the molarity of \(\mathrm{CrO}_{4}^{2-}\) ions, which is: \[ [\mathrm{Tl}^{+}] = 2 \times [\mathrm{CrO}_{4}^{2-}] = 2 \times 6.3 \times 10^{-5} \mathrm{~mol} \cdot \mathrm{L}^{-1} = 1.26 \times 10^{-4} \mathrm{~mol}\cdot\mathrm{L}^{-1}\]

Step 4: Calculate the Solubility Product Constant \(K_{\mathrm{sp}}\)

Substitute the concentrations of \(\mathrm{Tl}^{+}\) and \(\mathrm{CrO}_{4}^{2-}\) into the \(K_{\mathrm{sp}}\) expression: \[K_{\mathrm{sp}} = (1.26 \times 10^{-4})^{2} \times (6.3 \times 10^{-5}) \] Now calculate the value of \(K_{\mathrm{sp}}\): \[K_{\mathrm{sp}} = (1.26 \times 10^{-4})^{2} \times (6.3 \times 10^{-5}) \approx 1.00 \times 10^{-12}\]

Key Concepts

Molarity

Molarity is the measure of concentration of a solute in a solution. It's defined as the number of moles of the solute divided by the total volume of the solution in liters. The formula is represented as:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Understanding molarity is crucial when dealing with chemical reactions in solution because it allows us to relate amounts of reactants and products. In the context of our exercise, the molarity of chromate ions \(\mathrm{CrO}_{4}^{2-}\) tells us how much chromate is present in a given volume of the saturated solution of \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\), which is a key piece of information for calculating the solubility product constant.

Ksp Calculation

The solubility product constant, represented as \(K_{\mathrm{sp}}\), provides information about the solubility of a compound. It is calculated using the molar concentrations of the ions in a saturated solution at equilibrium. To calculate \(K_{\mathrm{sp}}\), each ionic concentration is raised to the power of its coefficient in the balanced dissolution equation and then multiplied together.
For our example, after establishing the molar concentrations of the ions involved, we plug these into the formula:
\[K_{\mathrm{sp}} = [\mathrm{Tl}^{+}]^{2}[\mathrm{CrO}_{4}^{2-}]\]
The calculation of \(K_{\mathrm{sp}}\) helps in predicting whether a precipitate will form in a particular solution and is crucial for chemists working with solutions at chemical equilibrium.

Chemical Equilibrium

Chemical equilibrium occurs when the rate of the forward reaction equals the rate of the reverse reaction, leading to a stable ratio of products and reactants. It doesn't mean the reactants and products are in equal concentrations, but rather that their concentrations do not change over time. In the equilibrium expression for \(K_{\mathrm{sp}}\), only the concentrations of dissolved ions are included, since the concentration of a pure solid, like our \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\), is constant and does not appear in the expression.
Understanding equilibrium is essential in predicting the behavior of a system under various conditions, such as changes in concentration or temperature, and it is a core concept in understanding solubility and precipitation.

Dissociation of Salts

The dissociation of salts into their constituent ions in water is an important chemical process for understanding solubility. This process is described by a dissolution equation that shows the salt dissociating into its ions. For instance, \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\) dissociates into \(\mathrm{Tl}^{+}\) and \(\mathrm{CrO}_{4}^{2-}\) ions.
In a saturated solution, the dissolved salt and the undissolved salt are in equilibrium, with the dissolved ions in the solution constant at a given temperature. The degree of dissociation, which varies for different salts, determines the salt's solubility in water. Salts like \(\mathrm{Tl}_{2}\mathrm{CrO}_{4}\), which do not fully dissociate in water, have a lower solubility characterized by a smaller \(K_{\mathrm{sp}}\), indicating a less soluble compound.