Question

The solubility product of \(\mathrm{AgCl}\) is \(1.5625 \times 10^{-10}\) at \(25^{\circ} \mathrm{C}\). Its solubility in grams per litre will be (a) \(143.5\) (b) 108 (c) \(1.57 \times 10^{-8}\) (d) \(1.79 \times 10^{-3}\)

Short answer

\(1.79 \times 10^{-3}\) g/L

Step-by-step solution

- Write the Dissociation Equation

Write the balanced equation for the dissociation of \(\mathrm{AgCl}\) into its ions in water: \[\mathrm{AgCl} \rightleftharpoons \mathrm{Ag^{+}} + \mathrm{Cl^{-}}\]

- Express the Solubility Product Constant

Express the solubility product constant (Ksp) for the dissociation of \(\mathrm{AgCl}\). Ksp is the product of the concentrations of the ions at equilibrium: \[\mathrm{Ksp} = [\mathrm{Ag^{+}}] [\mathrm{Cl^{-}}]\] where \[\mathrm{Ksp} = 1.5625 \times 10^{-10}\]

- Set up the Expression for Solubility

Let the solubility of \(\mathrm{AgCl}\) in water be \(s\) mol/L. At equilibrium, the concentration of \(\mathrm{Ag^{+}}\) and \(\mathrm{Cl^{-}}\) are both \(s\), since \(\mathrm{AgCl}\) dissociates into one mole of each per mole dissolved. So, \[\mathrm{Ksp} = s \times s = s^2\]

- Calculate the Solubility \(s\)

Using the value for Ksp, find the solubility \(s\): \[\mathrm{Ksp} = s^2 = 1.5625 \times 10^{-10}\] Solve for \(s\): \[\s = \sqrt{1.5625 \times 10^{-10}} = 1.25 \times 10^{-5}\] mol/L

- Convert Solubility to Grams per Litre

To find solubility in grams per litre, multiply the molar solubility by the molar mass of \(\mathrm{AgCl}\) (143.5 g/mol): \[\s \text{ in g/L } = (1.25 \times 10^{-5} \text{ mol/L}) \times (143.5 \text{ g/mol})\] \[\s \text{ in g/L } = 1.79 \times 10^{-3}\text{ g/L}\]

Key concepts

Ksp Calculation

Understanding how to calculate the solubility product constant (\r\( Ksp \)) is vital for predicting the solubility of ionic compounds in water. The \r\( Ksp \) is a mathematical representation of a saturated solution's ionic equilibrium, where a solid is in dynamic balance with its ions. It is determined by the product of the concentrations of the ions involved in the equilibrium, each raised to the power of its stoichiometric coefficient in the balanced dissolution equation.

Here's a simplified way to calculate \r\( Ksp \): First, you need the balanced dissolution equation of the compound. For instance, silver chloride (\r\( AgCl \)) dissociates into silver ions (\r\( Ag^+ \)) and chloride ions (\r\( Cl^- \)) in a one-to-one ratio. With this ratio, the concentrations of both ions at equilibrium are equal, and the \r\( Ksp \) expression becomes the concentration of one ion squared (\r\( [Ag^+][Cl^-] = s^2 \)) where \r\( s \) is the molar solubility. By knowing the \r\( Ksp \), you can solve for \r\( s \), which is the first step in determining the solubility of the compound in grams per litre.

Ionic Equilibrium

Ionic equilibrium pertains to a state of balance where the rate of the forward reaction (in which the solid ionic compound dissolves) equals the rate of the reverse reaction (in which ions combine to form the solid compound) in a saturated solution. This balance is crucial in understanding the dissolution and precipitation of ionic compounds.

The equilibrium concept can be extended to the solubility product. For the silver chloride example, when \r\( AgCl \) is dissolved in water, it will keep dissociating until the product of the ion concentrations equals the \r\( Ksp \). Beyond this point, any additional \r\( AgCl \) will not dissolve; instead, it'll start to precipitate. Therefore, the \r\( Ksp \) value is a measure of the solubility of the compound: the higher the \r\( Ksp \), the more soluble the compound. However, \r\( Ksp \) values are temperature-dependent, meaning that equilibrium concentrations can change with temperature.

Molar Solubility

Molar solubility is the number of moles of a solute that can be dissolved per litre of solution before the solution becomes saturated. It's an essential concept when dealing with the solubility of ionic compounds, which are often expressed in terms of how many moles can dissolve to form a saturated solution.

For example, let the molar solubility of \r\( AgCl \) be represented as \r\( s \). Here we infer that one mole of \r\( AgCl \) dissociates to form one mole each of \r\( Ag^+ \) and \r\( Cl^- \). Thus, at saturation, the concentrations of both ions are \r\( s \), and \r\( s^2 \) equals the \r\( Ksp \). By calculating \r\( s \) and knowing the molar mass of \r\( AgCl \), you can then find the solubility in grams per litre, which is \r\( s \) multiplied by the molar mass. This final conversion makes it easier to understand and measure solubility from a practical standpoint, such as when determining how much of a substance can be dissolved in a given amount of solvent in laboratory or industrial applications.