What is the molar solubility of AgIO₃ if K _sp = 3.0×10^{- 8}?

Question

Quizplus · Accepted Answer

The expression for the solubility product constant, Ksp, is:

Ksp = [Ag+][IO3-]

Let x be the molar solubility of AgIOS, meaning that [Ag+] = [IO3-] = x. Substituting this into the Ksp expression, we get:

Ksp = x2

We can solve for x by taking the square root of Ksp:

x = √(Ksp)

x = √(3.0×10-8)

x = 1.7×10-4 M

Therefore, the molar solubility of AgIOS is 1.7×10-4 M, which corresponds to choice B.

What Is the Molar Solubility of AgIO3 If K sp