What is the molar concentration of silver ion in a solution containing 1.3 × 10–4 M CrO42–, saturated with Ag2CrO4?
The Solubility Product Constant, Ksp for Ag2CrO4 is 9 × 10–12
(A) 1.3 × 10–16
(B) 7 × 10–16
(C) 9 × 10–12
(D) 2.6 × 10–4
(E) 7 × 10–3
This is a common ion worked the same way as the BaSO4 and Na2SO4 problem.
1.3x10^-4 is the answer to the question.
To find the molar concentration of silver ion in the solution, we need to use the solubility product constant (Ksp) and the concentration of chromate ion (CrO42–). Here's how you can calculate it:
Step 1: Write the balanced equation for the dissolution of Ag2CrO4.
Ag2CrO4(s) ⇌ 2Ag+(aq) + CrO42–(aq)
Step 2: Set up the Ksp expression using the balanced equation.
Ksp = [Ag+]^2 [CrO42–]
Step 3: Plug in the values given in the question.
Ksp = (x)^2 (1.3 × 10–4)
Ksp = 9 × 10–12
Step 4: Solve for x, which represents the molar concentration of Ag+.
(x)^2 = (9 × 10–12)/(1.3 × 10–4)
x^2 = 6.923 × 10^-8
x ≈ 8.32 × 10^-5
The molar concentration of silver ion (Ag+) in the solution is approximately 8.32 × 10^-5 M.
Since this value is not an option among the given choices, it seems that there might be an error in the question.