Question

Given $${K_{sp}}(AgI) = 8.5 \times {10^{ - 17}}$$. The solubility of AgI in 0.1M KI solution is
A. 0.1M
B. $$8.5 \times {10^{ - 16}}M$$
C. $$8.5 \times {10^{ - 17}}M$$
D. $$8.5 \times {10^{ - 15}}M$$

Answer 1

Hint: Solubility is defined as the ability of the substance or the maximum amount of substance that can dissolve in the solvent at a specific temperature. Solubility of the substance increases with its $${K_{sp}}$$ value. The solubility product is known as $${K_{sp}}$$ which is the equilibrium constant for a solid substance which gets dissolved in aqueous solution.

Complete step by step answer:
The equation of dissociation of KI is the following
Initially if we have 0.1 M of KI, then the ions of potassium in the solution will be 0.1M and of iodine will be 0.1M.
The equation of dissociation for AgI is the following
At equilibrium if we assume the solubility of silver ion and iodine ion in silver iodide solution to be ‘s’. but as the iodine ion is already present in solution with 0.1M so its total solubility will be (s+0.1).
Now if we make the expression for $${K_{sp}}$$that will be the product of silver ion and iodine ion,
$${K_{sp}} = [A{g^ + }][{I^ - }]$$
= s(s+0.1)
Now as the silver iodide salt is sparingly soluble salt so its molar solubility will be very low so we can assume that s<<<0.1, then we can ignore ‘s’ in (s+0.1), so$$s \approx 0.1$$
So the expression will be
$$8.5 \times {10^{ - 17}}$$=s(0.1)
$$s = \dfrac{{8.5 \times {{10}^{ - 17}}}}{{0.1}}$$
s= $$8.5 \times {10^{ - 16}}M$$

So, the correct answer is Option B.

Note: The solubility product depends upon lattice enthalpy of the salt and also on salvation enthalpy of ions in the solution. The energy is always released in this process so salvation enthalpy of ions is negative. When salt gets dissolved in solvent the intermolecular attraction between the particles of solute is overcome by the interactions between ions and solvent.