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QUESTION

# Solubility product of Hg2Cl2 at 25 C is 1.2*10^-18 mol^3 dm^-9. The concentration of Hg2+^2+ ions (in M) in a 0.040 M NaCl (aq) sol. saturated with Hg2Cl2 at 25 C is?

7.5x10^-16M

This calculation is solved using the common ion effect and .

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 = 7.5x10^-16M

To check if our assumption is valid substitute in the original unsimplified expression for and you will get 1.2x10^-18. So our simplification is valid.