Determine the molar solubility of AgBr in a solution containing 0.150 M NaBr. K_sp (AgBr)= 7.7 × 10^-13. A)8.8 × 10^-7 M B)3.9 × 10^-13M C)5.8 × 10^-5 M D)5.1 × 10^-12 M E)0.150 M

Question

Quizplus · Accepted Answer

We can write the solubility product expression for AgBr as:

Ksp = [Ag+][Br-]

Let the molar solubility of AgBr be x. Then, we can write:

[Ag+] = x
[Br-] = 0.150 (from the given concentration of NaBr)

Substituting these values into the Ksp expression, we get:

7.7 × 10^-13 = x × 0.150

Solving for x, we get:

x = 5.1 × 10^-12 M

Therefore, the molar solubility of AgBr in the given solution is 5.1 × 10^-12 M, which corresponds to answer choice D.

Determine the Molar Solubility of AgBr in a Solution Containing