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Chapter 16: Problem 78

Give an example to illustrate the general effect of complex ion formation on solubility.

Short Answer

Expert verified
The formation of complex ions increases the solubility of precipitates. In the example given, the introduction of ammonia (NH3) to a solution of silver chloride (AgCl) results in the formation of a complex ion \(Ag(NH3)2^+\), which in turn increases the solubility of the AgCl.

Step by step solution

01

Understanding the basic concept

We should understand that when a complex ion is formed, it increases the solubility of the precipitate. This is because forming a complex ion changes the position of equilibrium in line with Le Chatelier’s principle. A standard example would be the effect of ammonia on the solubility of silver chloride (AgCl).
02

Illustrating the process without complex ion formation

We can initially consider the solubility of AgCl in water without the presence of ammonia. The solubility equilibrium can be represented as follows: \[ AgCl(s) \leftrightarrow Ag^+(aq) + Cl^-(aq) \]The equilibrium constant expression for this dissolution process is \(K_{sp} = [Ag^+][Cl^-]\), where \(K_{sp}\) is the solubility product constant. AgCl is not very soluble in water, so \(K_{sp}\) is quite small.
03

Illustrating the process upon complex ion formation

Now consider the case when we add ammonia (NH3) to the solution. Ammonia reacts with \(Ag^+\) ion to form the complex ion \(Ag(NH3)2^+\). This can be represented by the equation:\[ Ag^+(aq) + 2NH3(aq) \leftrightarrow Ag(NH3)2^+(aq) \]The forward reaction demonstrates complex formation, which reduces the concentration of \(Ag^+\) ions in the solution. According to Le Chatelier’s principle, this causes the equilibrium in step 2 to shift to the right to compensate for the reduced \(Ag^+\) concentration, ultimately increasing the solubility of AgCl.
04

Summarize the effect of complex ion formation

The formation of complex ions increases the solubility of precipitates. This is because the reaction used to form the complex ion removes one of the products of the solubility equilibrium (in this case, \(Ag^+\) ions), causing the position of equilibrium to shift in order to increase the concentration of the removed species.

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Most popular questions from this chapter

The solubility product of \(\mathrm{PbBr}_{2}\) is \(8.9 \times 10^{-6}\). Determine the molar solubility (a) in pure water, (b) in \(0.20 M \mathrm{KBr}\) solution, (c) in \(0.20 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution.

If \(20.0 \mathrm{~mL}\) of \(0.10 \mathrm{MBa}\left(\mathrm{NO}_{3}\right)_{2}\) are added to \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3},\) will \(\mathrm{BaCO}_{3}\) precipitate?

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The \(p K_{a}\) of the indicator methyl orange is \(3.46 .\) Over what pH range does this indicator change from 90 percent HIn to 90 percent In \(^{-} ?\)
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