Refer to the Ksp values in Table 17.2 to calculate the molar solubility of each compound in pure water. a. AgBr

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Identify the Ksp value for AgBr from Table 17.2. Ksp (solubility product constant) represents the product of the concentrations of the ions in a saturated solution of a sparingly soluble salt.

Write the dissolution equation for AgBr. AgBr dissolves in water according to the equation: \( AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq) \).

Set up the expression for Ksp based on the dissolution equation. For AgBr, the Ksp expression is \( Ksp = [Ag^+][Br^-] \).

Assume the molar solubility of AgBr is 's'. This means that if 's' moles of AgBr dissolve, it will produce 's' moles of Ag+ and 's' moles of Br- in solution.

Substitute 's' into the Ksp expression and solve for 's'. The equation becomes \( Ksp = s \times s = s^2 \). Solving for 's' will give the molar solubility of AgBr in pure water.

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주요 개념

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Solubility Product Constant (Ksp)

The solubility product constant (Ksp) is an equilibrium constant that quantifies the solubility of a sparingly soluble ionic compound. It is defined as the product of the molar concentrations of the ions, each raised to the power of their coefficients in the balanced dissolution equation. For example, for AgBr, Ksp = [Ag+][Br-]. Understanding Ksp is essential for calculating the molar solubility of compounds in water.

Molar Solubility

Molar solubility refers to the maximum amount of a solute that can dissolve in a given volume of solvent at equilibrium, expressed in moles per liter (M). It is directly related to Ksp, as the molar solubility can be derived from the Ksp expression by substituting the concentrations of the ions in terms of the solubility. This concept is crucial for determining how much of a compound can dissolve in water.

Dissolution Equilibrium

Dissolution equilibrium describes the dynamic balance between the solid phase of a solute and its ions in solution. When a solid ionic compound like AgBr dissolves, it dissociates into its constituent ions, and the process reaches equilibrium when the rate of dissolution equals the rate of precipitation. Understanding this equilibrium is vital for calculating molar solubility and interpreting Ksp values.