The solubility of \(\mathrm{Ag}(\mathrm{I})\) in aqueous solutions containing different concentrations of \(\mathrm{Cl}^{-}\) is based on the following equilibria: $$ \begin{array}{lr} \mathrm{Ag}^{+}(a q)+\mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{AgCl}(s) & K_{\mathrm{sp}}=1.8 \times 10^{-10} \\ \mathrm{Ag}^{+}(a q)+2 \mathrm{Cl}^{-}(a q) & \rightleftharpoons \mathrm{AgCl}_{2}^{-}(a q) & K_{\mathrm{f}}=1.8 \times 10^{5} \end{array} $$ When solid \(\mathrm{AgCl}\) is shaken with a solution containing \(\mathrm{Cl}^{-}, \mathrm{Ag}(\mathrm{I})\) is present as both \(\mathrm{Ag}^{+}\) and \(\mathrm{AgCl}_{2}^{-}\). The solubility of \(\mathrm{AgCl}\) is the sum of the concentrations of \(\mathrm{Ag}^{+}\) and \(\mathrm{AgCl}_{2}^{-}\) (a) Show that \(\left[\mathrm{Ag}^{+}\right]\) in solution is given by $$ \left[\mathrm{Ag}^{+}\right]=1.8 \times 10^{-10} /\left[\mathrm{Cl}^{-}\right] $$ and that \(\left[\mathrm{AgCl}_{2}^{-}\right]\) in solution is given by $$ \left[\mathrm{AgCl}_{2}^{-}\right]=\left(3.2 \times 10^{-5}\right)\left(\left[\mathrm{Cl}^{-}\right]\right) $$ (b) Find the \(\left[\mathrm{Cl}^{-}\right]\) at which \(\left[\mathrm{Ag}^{+}\right]=\left[\mathrm{AgCl}_{2}^{-}\right]\). (c) Explain the shape of a plot of \(\mathrm{AgCl}\) solubility vs. \(\left[\mathrm{Cl}^{-}\right]\). (d) Find the solubility of \(\mathrm{AgCl}\) at the \(\left[\mathrm{Cl}^{-}\right]\) of part \((\mathrm{b}),\) which is the minimum solubility of \(\mathrm{AgCl}\) in the presence of \(\mathrm{Cl}^{-}\).

Short Answer

Expert verified
[Cl^-] = 7.5 \times 10^{-4} M. Solubility of AgCl is 4.8 \times 10^{-7} M.

Step by step solution

01

Determine \([Ag^+]\) expression

The solubility product constant expression (Ksp) for \({AgCl}\) is given by \[K_{sp} = [Ag^+][Cl^-]\]. Rearrange this expression to solve for \[ [Ag^+] \] in terms of \[ [Cl^-] \]: \[ [Ag^+] = \frac{K_{sp}}{[Cl^-]} \]. Substitute the given \[ K_{sp} = 1.8 \times 10^{-10} \]: \[ [Ag^+] = \frac{1.8 \times 10^{-10}}{[Cl^-]} \].
02

Determine \([AgCl_2^-]\) expression

The formation constant (Kf) for \[AgCl_2^- \] is given by \[ K_f = \frac{[AgCl_2^-]}{[Ag^+][Cl^-]^2} \]. Rearrange this expression to solve for \[ [AgCl_2^-] \] in terms of \[ [Ag^+] \] and \[ [Cl^-] \]: \[ [AgCl_2^-] = K_f [Ag^+][Cl^-]^2 \]. Use the expression from Step 1 for \[ [Ag^+] \]: \[ [AgCl_2^-] = K_f \left(\frac{1.8 \times 10^{-10}}{[Cl^-]}\right)[Cl^-]^2 = K_f \left(1.8 \times 10^{-10}\right)[Cl^-] \]. Substitute the given \[ K_f = 1.8 \times 10^5 \]: \[ [AgCl_2^-] = \left(1.8 \times 10^5\right) \left(1.8 \times 10^{-10}\right) [Cl^-] = 3.2 \times 10^{-5} [Cl^-] \].
03

Find \([Cl^-]\) for equal concentrations

Set the expressions for \[ [Ag^+] \] and \[ [AgCl_2^-] \] equal to each other to find the concentration of \[ [Cl^-] \] where they are equal: \[ \frac{1.8 \times 10^{-10}}{[Cl^-]} = 3.2 \times 10^{-5} [Cl^-] \]. Multiply both sides by \[ [Cl^-] \]: \[ 1.8 \times 10^{-10} = 3.2 \times 10^{-5} [Cl^-]^2 \]. Divide both sides by \[ 3.2 \times 10^{-5} \]: \[ [Cl^-]^2 = \frac{1.8 \times 10^{-10}}{3.2 \times 10^{-5}} = 5.625 \times 10^{-6} \]. Take the square root of both sides: \[ [Cl^-] = \sqrt{5.625 \times 10^{-6}} = 7.5 \times 10^{-4} \].
04

Explain solubility plot shape

The solubility of \[ AgCl \] vs. \[ [Cl^-] \] has a minimum at a particular \[ [Cl^-] \]. For low \[ [Cl^-] \], increasing \[ [Cl^-] \] will initially increase the solubility due to the formation of \[ {AgCl_2^-} \]. At high \[ [Cl^-] \], the formation of complex ions \[ {AgCl_2^-} \] becomes predominant, further increasing solubility.
05

Find solubility at minimum point

Use the expressions for \[ [Ag^+] \] and \[ [AgCl_2^-] \] found in steps 1 and 2. At \[ [Cl^-] = 7.5 \times 10^{-4} \], \[ [Ag^+] = 2.4 \times 10^{-7} \] and \[ [AgCl_2^-] = 2.4\times 10^{-7} \]. The total solubility is then \[ [Ag^+] + [AgCl_2^-] = 2.4 \times 10^{-7} + 2.4 \times 10^{-7} = 4.8 \times 10^{-7} \].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

solubility product constant
The solubility product constant, denoted as \(\text{K}_{\rm sp}\), is essential in understanding the solubility of sparingly soluble salts. Here, for \(\text{AgCl}\), it's given by \[ \text{K}_{\rm sp} = [\text{Ag}^+ ][\text{Cl}^{-}] \]. This equation states that the product of the ion concentrations of \(\text{Ag}^+\) and \(\text{Cl}^-\) in a saturated solution is constant at a given temperature. When solving for \(\text{Ag}^+\), we rearrange it to \[ \text{[Ag}^+]= \frac{\text{K}_{\rm sp}}{[\text{Cl}^- ]} \]. By substituting \(\text{K}_{\rm sp} = 1.8 \times 10^{-10}\), the concentration of \(\text{Ag}^+\) is inversely proportional to the concentration of \(\text{Cl}^−\). This explains why increasing \([\text{Cl}^-]\) decreases \([\text{Ag}^+]\). This inverse relationship forms the basis for understanding solubility in diverse chemical environments.
formation constant
The formation constant, denoted as \(\text{K}_f\), describes the equilibrium between free ions and complex ions. For the exercise, the formation of \(\text{AgCl}_2^−\) from \[ \text{Ag}^+ + 2\text{Cl}^{−} \rightleftharpoons \text{AgCl}_{2}^- \], is given by the expression \[ \text{K}_f = \frac{\text{[AgCl}_2^-]}{\text{[Ag}^+] \text{[Cl}^{-}]^2} \]. Rearranging it, we find the concentration of \(\text{AgCl}_2^-\) as \[ \text{[AgCl}_2^-] = \text{K}_f \text{[Ag}^+] \text{[Cl}^-]^2 \]. By substituting the value for \(\text{K}_f = 1.8 \times 10^5\), we get \[ \text{[AgCl}_2^-] = (1.8 \times 10^5)(1.8 \times 10^{-10})[\text{Cl}^-] = 3.2 \times 10^{-5}[\text{Cl}^-] \]. Formation constants are crucial for predicting the stability of complex ions in solutions.
complex ion formation
Complex ion formation is when metal ions bind with multiple ligands to form a more complex species, often leading to increased solubility. In our exercise, \(\text{AgCl}_{2}^-\) is the complex ion formed when \(\text{Ag}^+\) binds with two chloride ions. Examining the related reactions: \[ \text{Ag}^+ + \text{Cl}^- \rightleftharpoons \text{AgCl}(s) \] with \[ \text{K}_{\rm sp}=1.8 \times 10^{-10} \] and \[ \text{Ag}^+ + 2 \text{Cl}^- \rightleftharpoons \text{AgCl}_{2}^- \] with \[ \text{K}_f=1.8 \times 10^{5} \], complex ions significantly affect the solubility of salts. As \([\text{Cl}^-]\) increases, forming complex ions like \(\text{AgCl}_2^-\) makes more ions available in the solution, altering the solubility dynamics.
solubility calculation
To calculate solubility, we add the concentrations of all species containing the metal ion. For \(\text{AgCl}\), it includes \([\text{Ag}^+]\) and \[ \text{AgCl}_2^-\]. Solubility (\text{S}) therefore is \[ S = \text{[Ag}^+] + [\text{AgCl}_2^-] \]. In the exercise, at \[ [\text{Cl}^-] = 7.5 \times 10^{-4} M \], \[ \text{[Ag}^+] = \frac{1.8 \times 10^{-10}}{7.5 \times 10^{-4}} = 2.4 \times 10^{-7} M \] and \[ \text{[AgCl}_2^-] = 3.2 \times 10^{-5} \times 7.5 \times 10^{-4} = 2.4 \times 10^{-7} M \]. Adding these yields a total solubility S of \[ 2.4 \times 10^{-7} + 2.4 \times 10^{-7} = 4.8 \times 10^{-7} M \]. By understanding the contribution of each species, you ensure accurate solubility calculation.
chemical equilibrium
Chemical equilibrium is critical when dealing with solubility equilibria. It dictates that at equilibrium, the rates of forward and reverse reactions are equal, which ensures constant concentrations of reactants and products over time. In the context of the solubility of \(\text{AgCl}\), two equilibria are at play: the dissolution \[ \text{Ag}^+ (aq) + \text{Cl}^− (aq) \rightleftharpoons \text{AgCl}(s) \] and \[ \text{Ag}^+ (aq) + 2\text{Cl}^− (aq) \rightleftharpoons \text{AgCl}_2^{−} (aq) \]. The equilibrium constants \(\text{K}_{\rm sp} \) and \(\text{K}_{f} \) respectively describe these equilibria, ensuring that the overall concentrations of species involved remain steady. Manipulating conditions like concentration of \[ \text{Cl}^- \] thereby shifts the equilibrium, impacting solubility and the formation of complex ions.

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Most popular questions from this chapter

How do the acid and base components of a buffer function? Why are they often the conjugate acid-base pair of a weak acid?

Human blood contains one buffer system based on phosphate species and one based on carbonate species. Assuming that blood has a normal \(\mathrm{pH}\) of \(7.4,\) what are the principal phosphate and carbonate species present? What is the ratio of the two phosphate species? (In the presence of the dissolved ions and other species in blood, \(K_{\mathrm{a} 1}\) of \(\mathrm{H}_{3} \mathrm{PO}_{4}=1.3 \times 10^{-2}, K_{\mathrm{a} 2}=2.3 \times 10^{-7},\) and \(K_{\mathrm{a} 3}=6 \times 10^{-12} ; K_{\mathrm{a} 1}\) of \(\mathrm{H}_{2} \mathrm{CO}_{3}=8 \times 10^{-7}\) and \(\left.K_{\mathrm{a} 2}=1.6 \times 10^{-10} .\right)\)

Find the \(\mathrm{pH}\) of a buffer that consists of \(0.45 \mathrm{M} \mathrm{HCOOH}\) and \(0.63 M \mathrm{HCOONa}\left(\mathrm{p} K_{\mathrm{a}}\right.\) of \(\left.\mathrm{HCOOH}=3.74\right)\)

Tooth enamel consists of hydroxyapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) \(\left(K_{\mathrm{sp}}=6.8 \times 10^{-37}\right) .\) Fluoride ion added to drinking water reacts with \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) to form the more tooth decay-resistant fluorapatite, \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\left(K_{\mathrm{sp}}=1.0 \times 10^{-60}\right) .\) Fluoridated water has dramatically decreased cavities among children. Calculate the solubility of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{OH}\) and of \(\mathrm{Ca}_{5}\left(\mathrm{PO}_{4}\right)_{3} \mathrm{~F}\) in water.

Which compound in each pair is more soluble in water? (a) Strontium sulfate or barium chromate (b) Calcium carbonate or copper(II) carbonate (c) Barium iodate or silver chromate

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