Problem 76
Question
Solutions of sodium thiosulfate are used to dissolve unexposed \(\operatorname{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for black-and-white film. What mass of \(\mathrm{AgBr}\) can dissolve in \(1.00 \mathrm{L}\) of \(0.500 M\) \(\mathrm{Na}_{2} \mathrm{S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}^{2-}\) to form a complex ion: $$\begin{aligned} \mathrm{Ag}^{+}(a q)+2 \mathrm{S}_{2} \mathrm{O}_{3}^{2-}(a q) & \rightleftharpoons \mathrm{Ag}\left(\mathrm{S}_{2} \mathrm{O}_{3}\right)_{2}^{3-}(a q) \\ K &=2.9 \times 10^{13} \end{aligned}$$
Step-by-Step Solution
Verified Answer
In 1.00 L of 0.500 M Na$_2$S$_2$O$_3$, 2420.48 g of AgBr can dissolve.
1Step 1: Write the expressions of Ksp and K
First, write the expression for the solubility product constant (Ksp) for AgBr:
\[K_{sp} = [\mathrm{Ag}^+] [\mathrm{Br}^-]\]
Next, write the expression for the equilibrium constant (K) for the formation of the complex ion:
\[K = \frac{[\mathrm{Ag}(S_2O_3)_2^{3-}]}{[\mathrm{Ag}^+][\mathrm{S_2O_3}^{2-}]^2}\]
2Step 2: Calculate the concentration of Ag+ ions at equilibrium
The concentration of Ag+ ions at equilibrium can be found by solving for [Ag+] in the expressions for K and Ksp:
Rearrange the expression for K to get the concentration of Ag+ ions:
\[[\mathrm{Ag}^+] = \frac{[\mathrm{Ag}(S_2O_3)_2^{3-}]}{K[\mathrm{S_2O_3}]^2}\]
We know that the complex ion will form from Ag+ produced from the dissolution of the AgBr, so the concentration of the complex ion is equal to the concentration of Ag+ ions at equilibrium:
\[[\mathrm{Ag}(S_2O_3)_2^{3-}] = [\mathrm{Ag}^+]\]
Now, we can plug in the values for K and the concentration of S2O3^2- (from the given 0.500 M Na2S2O3 solution) into the expression for [Ag+]:
\[[\mathrm{Ag}^+] = \frac{[\mathrm{Ag}^+]}{(2.9 \times 10^{13})(0.500)^2}\]
Solve for [Ag+]:
\[[\mathrm{Ag}^+] = 3.88 \times 10^{-14} \, \text{M}\]
3Step 3: Calculate the mass of AgBr that can dissolve
Now that we have the concentration of Ag+ ions at equilibrium, we can use the Ksp expression to determine the concentration of Br^- ions at equilibrium:
\[[\mathrm{Br}^-] = \frac{K_{sp}}{[\mathrm{Ag}^+]}\]
Plug in values for Ksp and [Ag+]:
\[[\mathrm{Br}^-] = \frac{5.0 \times 10^{-13}}{3.88 \times 10^{-14}}\]
Solve for [Br^-]:
\[[\mathrm{Br}^-] = 12.89 \, \text{M}\]
Since the molar ratio between AgBr and Br^- ions is 1:1, the concentration of dissolved AgBr is the same as the concentration of Br^- ions. To find the mass of dissolved AgBr, multiply the concentration by the volume of the solution and the molar mass of AgBr:
\[\text{Mass of AgBr} = [\mathrm{Br}^-] \times \text{Volume} \times \text{Molar mass of AgBr}\]
Plug in values:
\[\text{Mass of AgBr} = (12.89 \, \text{M}) (1.00 \, \text{L}) (187.77 \, \frac{\text{g}}{\text{mol}})\]
Calculate the mass of AgBr:
\[\text{Mass of AgBr} = 2420.48 \, \text{g}\]
In 1.00 L of 0.500 M Na2S2O3, 2420.48 g of AgBr can dissolve.
Key Concepts
Complex Ion FormationEquilibrium ConstantChemical EquilibriumDissolution Process
Complex Ion Formation
Complex ion formation refers to a process where a central metal ion combines with ligands to form a complex ion. In the given exercise, the silver ion (\(\mathrm{Ag}^+\)) reacts with thiosulfate ions (\(\mathrm{S}_{2}\mathrm{O}_{3}^{2-}\)) to produce the complex ion \(\mathrm{Ag}(\mathrm{S}_{2} \mathrm{O}_{3})_{2}^{3-}\). This is a significant step since the formation of the complex ion impacts the solubility of \(\mathrm{AgBr}\) in the solution.
The tendency of silver ions to form such tightly bound complexes with thiosulfate increases their solubility. This is used in the photography industry, where unexposed silver bromide is dissolved using thiosulfate-containing solutions. Here, the concept is that forming a stable complex discourages the reverse reaction, enhancing the complete dissolution of \(\mathrm{AgBr}\).
Understanding complex ion formation helps explain why certain reactions proceed more efficiently by influencing the position of equilibrium and increasing or decreasing the concentration of particular ions in solution.
The tendency of silver ions to form such tightly bound complexes with thiosulfate increases their solubility. This is used in the photography industry, where unexposed silver bromide is dissolved using thiosulfate-containing solutions. Here, the concept is that forming a stable complex discourages the reverse reaction, enhancing the complete dissolution of \(\mathrm{AgBr}\).
Understanding complex ion formation helps explain why certain reactions proceed more efficiently by influencing the position of equilibrium and increasing or decreasing the concentration of particular ions in solution.
Equilibrium Constant
The equilibrium constant (\(K\)) is a vital aspect in understanding chemical reactions at equilibrium. It provides a quantitative measure of the position of equilibrium for a reversible reaction. In the context of this exercise, the equilibrium constant for the formation of the complex ion \(\mathrm{Ag}(\mathrm{S}_{2} \mathrm{O}_{3})_{2}^{3-}\) is extremely high, \(2.9 \times 10^{13}\).
This large value signifies that the reaction heavily favors the formation of the complex ion over the separated ions. This makes the complex ion very stable, effectively shifting the equilibrium towards the product side of the equation. The evaluation of \(K\) thus helps predict the extent of reaction or dissolution that will occur in a given chemical environment.
By understanding how to express and apply the equilibrium constant, students can better predict how large or small changes in concentration can affect the entire equilibrium system.
This large value signifies that the reaction heavily favors the formation of the complex ion over the separated ions. This makes the complex ion very stable, effectively shifting the equilibrium towards the product side of the equation. The evaluation of \(K\) thus helps predict the extent of reaction or dissolution that will occur in a given chemical environment.
By understanding how to express and apply the equilibrium constant, students can better predict how large or small changes in concentration can affect the entire equilibrium system.
Chemical Equilibrium
Chemical equilibrium refers to the state in which both the forward and reverse reactions occur at the same rate, leading to no observable changes in the concentrations of the reactants and products. In a dynamic system, such as the one described in this exercise with silver bromide and thiosulfate ions, chemical equilibrium dictates the extent to which \(\mathrm{AgBr}\) dissolves.
At equilibrium, the concentrations of reactants and products remain constant, although both reactions continue to occur. For the system here, equilibrium is achieved between the dissolution of \(\mathrm{AgBr}\) and formation of the \(\mathrm{Ag}(\mathrm{S}_{2} \mathrm{O}_{3})_{2}^{3-}\) complex ion.
This balance allows us to use measurable parameters, like the solubility product constant (\(K_{sp}\)) and \(K\), to predict concentrations of dissolved ions at equilibrium.
At equilibrium, the concentrations of reactants and products remain constant, although both reactions continue to occur. For the system here, equilibrium is achieved between the dissolution of \(\mathrm{AgBr}\) and formation of the \(\mathrm{Ag}(\mathrm{S}_{2} \mathrm{O}_{3})_{2}^{3-}\) complex ion.
This balance allows us to use measurable parameters, like the solubility product constant (\(K_{sp}\)) and \(K\), to predict concentrations of dissolved ions at equilibrium.
Dissolution Process
The dissolution process is essential to understanding how substances dissolve in solvents, involving breaking ionic bonds and forming new interactions with the solvent. The solubility of a compound like silver bromide (\(\mathrm{AgBr}\)) hinges on this process. Initially, solubility depends on \(K_{sp}\), the solubility product constant, which represents the maximum amount of \(\mathrm{AgBr}\) that can dissolve in water without forming more solid.
When complex ion formation enters the scene, it affects the dissolution process. The creation of a stable complex ion allows \(\mathrm{AgBr}\) to dissolve further than its usual solubility limit, because \(\mathrm{Ag}^+\) ions are effectively removed from the solution as they form the complex ion. This shift results in a greater amount of \(\mathrm{AgBr}\) dissolving, as can be calculated using the equilibrium constant expressions to find the necessary ion concentrations.
When complex ion formation enters the scene, it affects the dissolution process. The creation of a stable complex ion allows \(\mathrm{AgBr}\) to dissolve further than its usual solubility limit, because \(\mathrm{Ag}^+\) ions are effectively removed from the solution as they form the complex ion. This shift results in a greater amount of \(\mathrm{AgBr}\) dissolving, as can be calculated using the equilibrium constant expressions to find the necessary ion concentrations.
- The equilibrium state ensures consistent saturation levels for prolonged dissolution.
- Complex ions lower the effective concentration of dissociating ions, enhancing dissolution.
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