What is the Ksp of AgCl if the solubility of AgCl in water is 1.0 x 10^-5 M?
Practice Questions
1 question
Q1
What is the Ksp of AgCl if the solubility of AgCl in water is 1.0 x 10^-5 M?
1.0 x 10^-10
1.0 x 10^-5
1.0 x 10^-15
1.0 x 10^-20
Ksp = [Ag+][Cl-] = (1.0 x 10^-5)(1.0 x 10^-5) = 1.0 x 10^-10.
Questions & Step-by-step Solutions
1 item
Q
Q: What is the Ksp of AgCl if the solubility of AgCl in water is 1.0 x 10^-5 M?
Solution: Ksp = [Ag+][Cl-] = (1.0 x 10^-5)(1.0 x 10^-5) = 1.0 x 10^-10.
Steps: 7
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 (silver chloride) dissociates in water into Ag+ (silver ions) and Cl- (chloride ions).
Step 3: Write the dissociation equation: AgCl (s) ⇌ Ag+ (aq) + Cl- (aq).
Step 4: Note that the solubility of AgCl is given as 1.0 x 10^-5 M. This means that in a saturated solution, the concentration of Ag+ is 1.0 x 10^-5 M and the concentration of Cl- is also 1.0 x 10^-5 M.
Step 5: Use the Ksp formula: Ksp = [Ag+][Cl-].
Step 6: Substitute the values into the formula: Ksp = (1.0 x 10^-5)(1.0 x 10^-5).
