Question

Solutions of sodium thiosulfate are used to dissolve unexposed \(\mathrm{AgBr}\left(K_{\mathrm{sp}}=5.0 \times 10^{-13}\right)\) in the developing process for blackand-white film. What mass of AgBr can dissolve in \(1.00 \mathrm{~L}\) of \(0.500 \mathrm{M} \mathrm{Na}_{2} \mathrm{~S}_{2} \mathrm{O}_{3} ? \mathrm{Ag}^{+}\) reacts with \(\mathrm{S}_{2} \mathrm{O}_{3}{ }^{2-}\) to form a complex ion:

Short answer

The mass of AgBr that can dissolve in 1.00 L of 0.500 M Na2S2O3 solution is approximately \(3.32 \times 10^{-5} g\).

Step-by-step solution

Write the balanced chemical equation

The overall reaction can be written as follows: \(AgBr(s) \rightleftharpoons Ag^+(aq) + Br^-(aq)\) Since the deep-reaching complex ion is: \(Ag^+(aq) + 2S_2O_3^{2-}(aq) \rightarrow [Ag(S_2O_3)_2]^{3-}(aq)\)

Find the concentration of Ag+ and Br- ions using Ksp

We will use the solubility product constant (Ksp) to find the concentration of Ag+ and Br- ions at equilibrium. The Ksp expression for AgBr is: \(K_{sp} = [Ag^+][Br^-]\) We know that Ksp = 5.0 x 10^(-13). Let x be the concentration of Ag+ and Br- ions at equilibrium. Therefore: \(5.0 \times 10^{-13} = x^2\) Now, solve for x: \(x = \sqrt{5.0 \times 10^{-13}}\) \(x = 7.07 \times 10^{-7} M\) Hence, the concentration of Ag+ and Br- ions at equilibrium is \(7.07 \times 10^{-7} M\).

Use stoichiometry to find the final concentration of Ag+ and Br- ions

The initial concentration of Na2S2O3 is given as 0.500 M. In the reaction, one mole of Ag+ reacts with two moles of S2O3^2- to form the complex ion. Using stoichiometry, let y be the moles of Ag+ that reacted with S2O3^2-: \(\frac{y}{7.07 \times 10^{-7}} = \frac{2 \times y}{0.5}\) Solving for y, we get: \(y = 1.77 \times 10^{-7} M\)

Use the final concentrations to find the mass of AgBr

Now that we have found the concentration of Ag+ ions that reacted with Na2S2O3, we can calculate the mass of AgBr that can dissolve in the solution. Mass of AgBr = moles of Ag+ ions x molar mass of AgBr Molar mass of AgBr = \(107.87 g/mol (Ag)\) + \(79.9 g/mol (Br)\) = 187.77 g/mol Mass of AgBr = \((1.77 \times 10^{-7} mol) \times (187.77 g/mol)\) Mass of AgBr = \(3.32 \times 10^{-5} g\) Therefore, the mass of AgBr that can dissolve in 1.00 L of 0.500 M Na2S2O3 solution is approximately \(3.32 \times 10^{-5} g\).

Key concepts

Solubility Product Constant (Ksp)

The solubility product constant, often abbreviated as Ksp, is a crucial concept when dealing with sparingly soluble salts. In the context of sodium thiosulfate dissolving silver bromide (AgBr), Ksp helps us understand to what extent AgBr dissolves in water.

Solubility product is represented by the expression:

• \(K_{sp} = [Ag^+][Br^-]\)

This equation tells us that the product of the molar concentrations of the ions in solution, each raised to the power of their coefficients in the balanced equation, equals Ksp. For AgBr, the low Ksp value of \(5.0 \times 10^{-13}\) indicates that only a tiny amount dissolves in water.Understanding the Ksp allows us to calculate specific concentrations of ions in equilibrium. In this case, solving for the concentrations of \([Ag^+]\) and \([Br^-]\) tells us how much AgBr can be dissolved under given conditions.

Complex Ion Formation

When silver ions (Ag^+) react with thiosulfate ions (S_2O_3^{2-}), they form a complex ion. This formation is essential in increasing the solubility of sparingly soluble salts like AgBr.

In our reaction, the complex ion is:

• \(Ag^+(aq) + 2S_2O_3^{2-}(aq) \rightarrow [Ag(S_2O_3)_2]^{3-}(aq)\)

Here, one silver ion combines with two thiosulfate ions to form a negatively charged complex ion. This process is crucial because it reduces the concentration of free Ag^+ in solution.When the equilibrium shifts to the right due to complex ion formation, it effectively pulls more of the AgBr into solution. The complex ion formation thus aids in dissolving more AgBr than would be possible if only considering the Ksp alone.

Stoichiometry in Chemistry

Stoichiometry is a fundamental aspect of chemistry that involves using ratios from balanced chemical equations. For the dissolution of AgBr in sodium thiosulfate solution, stoichiometry helps us understand how many moles of reactants and products interact.

In the overall reaction of:

• \(Ag^+(aq) + 2S_2O_3^{2-}(aq) \rightarrow [Ag(S_2O_3)_2]^{3-}(aq)\)

We see that one mole of Ag^+ reacts with two moles of thiosulfate to form a complex ion. By applying stoichiometric principles, we calculate the final concentrations of ions in solution. Given an initial 0.500 M concentration of Na2S2O3, we used the stoichiometry ratio to determine how much Ag^+ is required.This calculation shows the relationship between reactants and helps us understand the extent of the complex ion formation, which plays a significant role in dissolving more AgBr.

Chemical Equilibrium Calculations

Understanding chemical equilibrium is vital in calculating how reactions proceed and to what extent. Equilibrium calculations allow us to predict the concentrations of reactants and products over time.

For the given system involving AgBr and sodium thiosulfate, Ksp provides the equilibrium condition for AgBr dissociation. We calculate initial concentrations using Ksp, then factor in complex ion formation to reassess those values. In this exercise, equilibrium calculations helped identify the final concentration of Ag+ addressed in stoichiometry.

• Initial concentrations were found using Ksp: \(x = 7.07 \times 10^{-7} M\)
• Complex ion changes the equilibrium dynamics, pulling more AgBr into solution.

These calculations ultimately inform our understanding of how much AgBr the solution can dissolve, accounting for all equilibrium-changing reactions.