Calculate the molar solubility of AgCl in a 1.0 M NH3 solution. Strategy AgCl is only slightly soluble in water: AgCl(s) ⇌ Ag^+(aq) + Cl^−(aq) The Ag^+ ions form a complex ion with NH3 (see Table 16.4):Ag^+(aq)+2NH3(aq)⇌ Ag(NH3)2^+ Combining these two equilibria will give the overall equilibrium

Question

alculate the molar solubility of AgCl in a 1.0 M [latex]NH_{3}[/latex] solution.

Strategy AgCl is only slightly soluble in water:

[latex]AgCl(s) \xrightleftharpoons[]{} Ag^{+}(aq) + Cl^{-}(aq)[/latex]

The [latex]Ag^{+}[/latex] ions form a complex ion with [latex]NH_{3}[/latex] (see Table 16.4):

[latex]Ag^{+}(aq)+2NH_{3}(aq) \xrightleftharpoons[]{} Ag(NH_{3})_{2}^{+}[/latex]

Combining these two equilibria will give the overall equilibrium for the process.

Accepted Answer

Step 1: Initially, the species in solution are [latex]Ag^{+}[/latex] and [latex]Cl^{-}[/latex] ions and [latex]NH_{3}[/latex]. The reaction between [latex]Ag^{+}[/latex] and [latex]NH_{3}[/latex] produces the complex ion [latex]Ag(NH_{3}) 2 +[/latex].

Step 2: The equilibrium reactions are

[latex]AgCl(s) \xrightleftharpoons[]{} Ag^{+}(aq)+Cl^{-}(aq)[/latex]

[latex]K_{sp}=[Ag^{+}][Cl^{-}]=1.6 imes 10^{-10}[/latex]

[latex]Ag^{+}(aq)+2NH_{3}(aq) \xrightleftharpoons[]{} Ag(NH_{3})^{+}_{2}(aq)[/latex]

[latex]K_{f}=\frac{[Ag(NH_{3})^{+}_{2}]}{[Ag^{+}][NH_{3}]^{2}} =1.5 imes 10^{7} [/latex]

[latex]\begin{matrix} \ Overall:&\begin{matrix} \hline \ &AgCl(s)+2NH_{3}(aq) \xrightleftharpoons[]{} Ag(NH_{3})^{+}_{2}(aq)+Cl^{-} \end{matrix} \end{matrix} [/latex]

The equilibrium constant K for the overall reaction is the product of the equilibrium constants of the individual reactions (see Section 14.2):

[latex]K=K_{sp}K_{f}=\frac{[Ag(NH_{3})^{+}_{2}][Cl^{-}]}{[NH_{3}]^{2}}[/latex]

[latex]=(1.6 imes 10^{-10} )(1.5 imes 10^{7} )[/latex]

[latex]=2.4 imes 10^{3} [/latex]

Let s be the molar solubility of AgCl (mol/L). We summarize the changes in concentrations that result from formation of the complex ion as follows:

[latex]\begin{matrix} & AgCl(s) & & + & 2NH_{3}(aq) & \xrightleftharpoons[]{} &Ag(NH_{3})_{2}^{+} (aq) &+&Cl^{-}(aq) \ Initial (M): & & &&1.0 &&0.0 &&0.0&&& \ Change (M): & -s & &&-2s&&+s &&+s& \ \hline Equilibrium (M):& &&&(1.0-2s)&&s &&s& \end{matrix} [/latex]

The formation constant for [latex]Ag(NH_{3})_{2}^{+}[/latex] is quite large, so most of the silver ions exist in the complexed form. In the absence of ammonia we have, at equilibrium, [latex][Ag^{+}][/latex] = [latex][Cl^{-}][/latex]. As a result of complex ion formation, however, we can write [latex][Ag(NH_{3}) 2 +][/latex] = [latex][Cl^{-}][/latex].

Step 3: [latex]K=\frac{(s)(s)}{(1.0-2s)^{2} } [/latex]

[latex]2.4 imes 10^{-3} =\frac{(s)(s)}{(1.0-2s)^{2} } [/latex]

Taking the square root of both sides, we obtain

[latex]0.049=\frac{(s)}{(1.0-2s) } [/latex]

[latex]s=0.045 M[/latex]

Step 4: At equilibrium, 0.045 mole of AgCl dissolves in 1 L of 1.0 M [latex]NH_{3}[/latex] solution.

Check The molar solubility of AgCl in pure water is [latex]1.3 imes 10^{-5}[/latex] M. Thus, the formation of the complex ion [latex]Ag(NH_{3})^{+}_{2}[/latex] enhances the solubility of AgCl (Figure 16.12).

Complex Ion	Equilibrium Expression	Formation Constant $(K_{f})$
$Ag(NH_{3})^{+}_{2}$	$Ag^{+}+2NH_{3} \xrightleftharpoons[]{} Ag(NH_{3})^{+}_{2}$	$1.5\times 10^{7}$
$Ag(CN)^{-}_{2}$	$Ag^{+}+2CN^{-} \xrightleftharpoons[]{} Ag(CN)^{-}_{2}$	$1.0\times 10^{21}$
$Cu(CN)^{2-}_{4}$	$Cu^{2+}+4CN^{-} \xrightleftharpoons[]{} Cu(CN)^{2-}_{4}$	$1.0\times 10^{25}$
$Cu(NH_{3})^{2+}_{4}$	$Cu^{2+}+4NH_{3} \xrightleftharpoons[]{} Cu(NH_{3})^{2+}_{4}$	$5.0\times 10^{13}$
$Cd(CN)^{2-}_{4}$	$Cd^{2+}+4CN^{-} \xrightleftharpoons[]{} Cd(CN)^{2-}_{4}$	$7.1\times 10^{16}$
$CdI^{2-}_{4}$	$Cd^{2+}+4I^{-} \xrightleftharpoons[]{} CdI^{2-}_{4}$	$2.0\times 10^{6}$
$HgCl^{2-}_{4}$	$Hg^{2+}+4Cl^{-} \xrightleftharpoons[]{} HgCl^{2-}_{4}$	$1.7\times 10^{16}$
$HgI^{2-}_{4}$	$Hg^{2+}+4I^{-} \xrightleftharpoons[]{} HgI^{2-}_{4}$	$2.0\times 10^{30}$
$Hg(CN)^{2-}_{4}$	$Hg^{2+}+4CN^{-} \xrightleftharpoons[]{} Hg(CN)^{2-}_{4}$	$2.5\times 10^{41}$
$Co(NH_{3})^{3+}_{6}$	$Co^{3+}+6NH_{3} \xrightleftharpoons[]{} Co(NH_{3})^{3+}_{6}$	$5.0\times 10^{31}$
$Zn(NH_{3})^{2+}_{4}$	$Zn^{2+}+4NH_{3} \xrightleftharpoons[]{} Zn(NH_{3})^{2+}_{4}$	$2.9\times 10^{9}$

Question 16.16: Calculate the molar solubility of AgCl in a 1.0 M NH3 soluti...

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