Question

The value of \[{K_{sp}}\]of \[HgC{l_2}\] at room temperature is \[4 \times {10^{ - 15}}\] . The concentration of \[C{l^ - }\] ion in its aqueous solution at saturation point is:
 \[
  A.\,\,\,1 \times {10^{ - 5}} \\
  B.\,\,\,\,2 \times {10^{ - 5}} \\
  C.\,\,\,\,2 \times {10^{ - 15}} \\
  D.\,\,\,\,8 \times {10^{ - 15}} \\
 \]

Answer 1

Hint: \[{K_{sp}}\] or the solubility product is basically the equilibrium between a solid and its corresponding ions in its solution. The value of \[{K_{sp}}\] varies according to the degree to which the compounds can dissolve or dissociate in a given aqueous solution.

Complete Step-by-Step Solution:
The given compound dissociates in the following manner:
\[HgC{l_2} \to H{g^{1 + }} + 2C{l^ - }\]
This means that for every one mole of mercury (Hg) that is dissolved in the aqueous solution of \[HgC{l_2}\], 2 moles of chlorine also dissolve. This indirectly interprets that the ratio of solubility of mercury to chlorine is 1 mole of mercury for every 2 moles of chlorine
\[
  {K_{sp}} = {[Hg]^1}{[Cl]^2} \\
  HgC{l_2} \to H{g^{1 + }} + 2C{l^ - } \\
    \\
 \]
\[{K_{sp}}\] can be represented as the product of the concentrations of the constituent elements each raised to their corresponding values of ratio of their dissociation in the aqueous solution. Mathematically representing the formula, we get:
\[{K_{sp}} = {[a]^x}{[b]^y}\]
\[{K_{sp}} = {[Hg]^1}{[Cl]^2}\]
\[{K_{sp}} = {[s]^1}{[4s]^2}\]
\[{K_{sp}} = 4{s^3}\]
Since the value of \[{K_{sp}}\]= \[4 \times {10^{ - 15}}\]
\[4 \times {10^{ - 15}} = 4{s^3}\]
Hence, \[s = {10^{ - 5}}\]
Now the solubility of Cl is 2s. Hence the total solubility of \[C{l^ - }\] is 2 \[ \times {10^{ - 5}}\]
Hence, the concentration of \[C{l^ - }\] ion in its aqueous solution at saturation point is 2 \[ \times {10^{ - 5}}\]

Hence, Option B is the correct option.

Note: the solubility product constant can vary due to the molecular size of the constituent atoms, the net applied pressure, temperature. It is also used for understanding the conditions under which a particular precipitate will be formed, and also helps to understand the common ion effect.