Question #d12fb

1 Answer
Jul 10, 2016



This calculation is solved using the common ion effect and Le Chatelier's principle.

In pure water the Mercury (1) chloride will have a low solubility. The concentration of the ions will be determined by the solubility product.

#Hg_2Cl_2 <=> Hg_2^(2+) + 2Cl^-#

In the presence of the NaCl solution Chloride ion will be the common ion. The presence of the chloride ion will cause the above equilibrium reaction to shift to the left in accordance with Le Chatelier's principle.

In the abscence of the Mercury compound we will have 0.04 M chloride ion in solution. When we add the Mercury compound we will have a fixed amount of #Hg_2^2+# ion. Lets assume this amount is expressed by the variable x M. Then the amount of chloride ion at equilibrium has to be 2x +0.04.

We know that the solubility product can be written as
#K_(sp) = [x][2x+0.04]^2#

We can expand out this into a trinomial and solve for x. However because we know that the solubility of this compound is very low we can make a simplifying assumption that x is very small. So we have
#K_(sp) = [x][0.04]^2#

We solve for x = 7.5x#10^-16#M

To check if our assumption is valid substitute in the original unsimplified expression for Ksp and you will get 1.2x#10^-18#. So our simplification is valid.