If the solubility product (Ksp) of AgCl is 1.77 x 10^-10, what is the molar solubility of AgCl?
Practice Questions
1 question
Q1
If the solubility product (Ksp) of AgCl is 1.77 x 10^-10, what is the molar solubility of AgCl?
1.77 x 10^-5 M
1.77 x 10^-10 M
1.77 x 10^-7 M
1.77 x 10^-3 M
Ksp = [Ag+][Cl-] = s^2, thus s = sqrt(1.77 x 10^-10) = 1.77 x 10^-5 M.
Questions & Step-by-step Solutions
1 item
Q
Q: If the solubility product (Ksp) of AgCl is 1.77 x 10^-10, what is the molar solubility of AgCl?
Solution: Ksp = [Ag+][Cl-] = s^2, thus s = sqrt(1.77 x 10^-10) = 1.77 x 10^-5 M.
Steps: 7
Step 1: Understand that Ksp (solubility product) is a constant that represents the solubility of a salt in water.
Step 2: Write the dissociation equation for AgCl: AgCl (s) ⇌ Ag+ (aq) + Cl- (aq).
Step 3: Recognize that when AgCl dissolves, it produces equal amounts of Ag+ and Cl- ions. If 's' is the molar solubility, then [Ag+] = s and [Cl-] = s.
Step 4: Write the expression for Ksp: Ksp = [Ag+][Cl-]. Since both concentrations are 's', we can write Ksp = s * s = s^2.
Step 5: Substitute the given Ksp value into the equation: 1.77 x 10^-10 = s^2.
Step 6: Solve for 's' by taking the square root of both sides: s = sqrt(1.77 x 10^-10).
Step 7: Calculate the square root: s = 1.77 x 10^-5 M.
