The solubility product constant of silver bromide is 5.0 × 10^-13. What is the molar solubility of this compound?

Question

Quizplus · Accepted Answer

The molar solubility of AgBr can be calculated using the expression:
Ksp = [Ag+][Br-]
where [Ag+] and [Br-] represent the molar concentrations of Ag+ and Br- ions in solution, respectively.
Since AgBr is a 1:1 electrolyte, its molar solubility can be expressed as x mol/L, so [Ag+] = x mol/L and [Br-] = x mol/L.
Substituting these values into the Ksp expression:
Ksp = x^2
x = (Ksp)^1/2 = ((5.0 × 10^-13)^(1/2))/1 mol/L = (50)^1/2 × 10^-7 mol/L
Therefore, the correct answer is option C, (50)^1/2 × 10^-7 mol/L.

The Solubility Product Constant of Silver Bromide Is 5