Question

The Solubility Product Constant for silver iodide is 1.5×10^-16.

If silver iodide is dissolved in water you can say that the equilibrium concentrations of silver and iodide ions are: ...

...A. High
...B. Moderate
...C. Low

The solubility of silver iodide in water is: ...
...A. High
...B. Moderate
...C. Low

Answer 1

Write down the ionization of silver iodide, AgI in water.

AgI (s) -------> Ag⁺ (aq) + I^- (aq)

K_sp = [Ag⁺][I^-]

====> 1.5*10^-16 = (x).(x) where x = molar concentration of Ag⁺ and I^- in water.

Therefore, 1.5*10^-16 = x²

====> x = 1.225*10^-8

The molar concentration of silver and iodide ions in water is 1.225*10^-8 M; obviously, the value is quite low and hence, the concentrations are low. This is expected; the value of the equilibrium constant is quite low and hence, the equilibrium favors the reactants over the products (ans).

The molar concentration of silver or iodide ion is equal to the molar solubility of silver iodide in water. This is again, low, since silver iodide mostly stays in the solid state.