Question: Calculate [HgCl₄^2-] in a solution prepared by adding 0.0200 mol of NaCl to 0.250 L of a 0.100 M HgCl₂ solution.

The reaction

\({K_f} = 1.1 \cdot {10^{16}}\)

• The concentration of HgCl₄solution is 0.100M
• The volume of a solution is 0.250 L
• 0.0200 mol of NaCl is added to a solution

Let us calculate the concentration of HgCl₄^2-

The number of moles of HgCl₄is

\(\begin{array}{*{20}{c}}{{n_{{\rm{HgC}}{{\rm{l}}_2}}} = 0.100{\rm{M}} \cdot 0.250{\rm{L}}}\\{ = 0.025{\rm{mol}}}\end{array}\)

• HgCl₂solution – 0.025 mol of HgCl₂ dissociates completely into 0.025 mol of Hg²⁺ and 0.05 mol of Cl^-
• After addition of 0.02 mol of NaCl (NaCl dissociates completely into 0.02 mol of Na⁺ and 0.02 mol of Cl^-), we have 0.025 mol of Hg²⁺and 0.07 mol of Cl^- (0.05 mol + 0.02 mol)

Since 1 mole of Hg²⁺reacts with 4 moles of Cl^-, the limiting reagent will be 0.07 mol of Cl^-

Since the value of\({K_f} = 1.1 \cdot {10^{16}}\) is very high, we will assume that the reaction will reach completion.

Four moles of Cl^- will produce one mole of HgCl₄^2-, and 0.07 mol of Cl^- will produce

\(\begin{array}{*{20}{c}}{{n_{{\rm{HgC}}{{\rm{l}}_4}^{2 - }}} = 0.07{\rm{molC}}{{\rm{l}}^ - }\frac{{4{\rm{molHgC}}{{\rm{l}}_4}^{2 - }}}{{1{\rm{molC}}{{\rm{l}}^ - }}}}\\{ = 1.75 \cdot {{10}^{ - 2}}{\rm{molHgC}}{{\rm{l}}_4}^{2 - }}\end{array}\)

And the concentration of HgCl₄^2-is

\(\left( {{\rm{HgCl}}_4^{2 - }} \right) = \frac{{1.75 \cdot {{10}^{ - 2}}{\rm{mol}}}}{{0.250{\rm{L}}}} = 0.07{\rm{M}}\)