Question

When equal volumes of the following solutions are mixed, precipitation of \[AgCl\] (${K_{sp}} = 1.8 \times {10^{ - 10}}$) will not occur only with:
A. ${10^{ - 4}}MA{g^ + }and{10^{ - 4}}MC{l^ - }$
B. ${10^{ - 3}}MA{g^ + }and{10^{ - 3}}MC{l^ - }$
C. ${10^{ - 6}}MA{g^ + }and{10^{ - 6}}MC{l^ - }$
D. ${10^{ - 2}}MA{g^ + }and{10^{ - 2}}MC{l^ - }$

Answer 1

Hint: Precipitation reactions usually occur when cations and anions in an aqueous solution combine to form an insoluble ionic solid called a precipitate. When equal volumes of some solutions are mixed, precipitation of \[AgCl\] will occur.

Complete step by step answer:
To find out if a solution will precipitate or dissolve, we calculate the ion product, $Q$ (same form as ${K_{sp}}$, but ion concentrations here are not necessarily equilibrium ion concentrations)
We compare $Q$ to ${K_{sp}}$
If $Q > {K_{sp}}$, the solution will be supersaturated, ion concentrations are greater than equilibrium concentrations, the reaction will proceed in reverse to reach equilibrium, precipitation will occur.
If $Q < {K_{sp}}$, the solution will be unsaturated, ion concentrations are less than equilibrium concentrations, the reaction will proceed forward to reach equilibrium, more solid will dissolve.
If $Q = {K_{sp}}$, the solution will be saturated, the solution is at equilibrium, ion concentrations are at equilibrium concentrations, no more solid will dissolve or even precipitate.
Given
${K_{sp}} = 1.8 \times {10^{ - 10}}$
When two solutions are mixed having equal volumes, the concentration will be reduced and it will become half the initial solution.
\[AgCl\]will precipitate only when $[A{g^ + }][C{l^ - }] > {K_{sp}}$
To assure this, we have to substitute values of the concentrations of silver and chloride ions given in the options above and if the value of the product of concentrations (the ion product $Q$) is greater than the value of ${K_{sp}}$, then precipitation of the solution will occur.
Given, in option (C)
As stated above when two solutions are mixed having equal volumes, the concentration will be reduced and it will become half the initial solution. Since, both silver and chloride have equal volumes and are mixed, the concentration of the silver and chloride ions will be:
$[A{g^ + }] = \dfrac{{{{10}^{ - 5}}}}{2}M$
$[C{l^ - }] = \dfrac{{{{10}^{ - 5}}}}{2}M$
Now, we calculate the product of the two concentrations called the ion product $Q$, to compare it to the value of the solubility constant ${K_{sp}}$
$[A{g^ + }][C{l^ - }] = \dfrac{{{{10}^{ - 5}}}}{2}M \times \dfrac{{{{10}^{ - 5}}}}{2}M$
On solving, we get: $[A{g^ + }][C{l^ - }] = \dfrac{{{{10}^{ - 10}}}}{4}M$
$ \Rightarrow Q = 2.5 \times {10^{ - 11}}M$
The value ${K_{sp}}$ is higher than the value of the ion product $Q$ in this case. Thus the solution will not precipitate.
$1.8 \times {10^{ - 10}} > 2.5 \times {10^{ - 11}}$
In options A, B, and D, on the calculation, the value of the ion product $Q$ is higher than the value of ${K_{sp}}$, thus all the three solutions will precipitate.

So, the correct answer is Option C.

Note: The solubility product constant is known as the equilibrium constant for the dissolution of a solid substance into an aqueous solution. It is usually denoted by the symbol ${K_{sp}}$
The solubility product is the kind of an equilibrium constant and its value depends on temperature. ${K_{sp}}$ usually increases with an increase in the temperature due to increased solubility.
Solubility is known as a property of a substance called solute to get dissolved in a solvent in order to form a solution. Some compounds can be highly soluble and may even absorb moisture from the atmosphere whereas others are highly insoluble.