What is the Ksp of AgCl if the solubility of AgCl in water is 1.33 x 10^-5 M? (2
Practice Questions
Q1
What is the Ksp of AgCl if the solubility of AgCl in water is 1.33 x 10^-5 M? (2023)
1.78 x 10^-10
1.33 x 10^-5
1.33 x 10^-10
1.78 x 10^-5
Questions & Step-by-Step Solutions
What is the Ksp of AgCl if the solubility of AgCl in water is 1.33 x 10^-5 M? (2023)
Step 1: Understand that Ksp (solubility product constant) is calculated using the concentrations of the ions in a saturated solution.
Step 2: Recognize that AgCl dissociates in water into Ag+ and Cl- ions. The equation is: AgCl (s) ⇌ Ag+ (aq) + Cl- (aq).
Step 3: Note that the solubility of AgCl is given as 1.33 x 10^-5 M. This means that in a saturated solution, the concentration of Ag+ ions is 1.33 x 10^-5 M and the concentration of Cl- ions is also 1.33 x 10^-5 M.
Step 4: Write the expression for Ksp: Ksp = [Ag+][Cl-].
Step 5: Substitute the values into the Ksp expression: Ksp = (1.33 x 10^-5)(1.33 x 10^-5).
Step 6: Calculate the Ksp: Ksp = 1.33 x 10^-5 * 1.33 x 10^-5 = 1.7689 x 10^-10, which can be rounded to 1.78 x 10^-10.
