If the solubility constant of silver iodide (AgI) is 8.5 × 10-17, what is the solubility of Ag+?
4.0 × 10^-13 M
2.0 × 10^-13 M
9.2 × 10^-9 M
4.6 × 10^-9 M
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To find the solubility of Ag+ given the solubility constant (Ksp) of silver iodide (AgI), we need to use the equation relating the solubility constant and the concentration of ions in the solution.
The equation for the solubility constant of AgI is as follows:
AgI ⇌ Ag+ + I-
Since the stoichiometric coefficient for Ag+ is 1 in the balanced chemical equation, the concentration of Ag+ is equal to the solubility (S) of AgI.
The expression for the solubility constant (Ksp) is given by:
Ksp = [Ag+][I-]
Given that the solubility constant (Ksp) of AgI is 8.5 × 10^-17, and we want to find the solubility of Ag+ (S), we can rearrange the equation to solve for S (Ag+ solubility):
S = [Ag+] = Ksp / [I-]
In this case, we need the concentration of I- to calculate the solubility of Ag+. However, the concentration of I- is not given in the question.
Without the concentration of I-, we cannot calculate the exact solubility of Ag+. Thus, none of the provided options (4.0 × 10^-13 M, 2.0 × 10^-13 M, 9.2 × 10^-9 M, 4.6 × 10^-9 M) can be determined without additional information.