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Chapter 16: Problem 70

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

Expert verified
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

01

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: \(Ag^+(aq) + 2S_2O_3^{2-}(aq) \rightarrow [Ag(S_2O_3)_2]^{3-}(aq)\)
02

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 \(7.07 \times 10^{-7} M\).
03

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: \(y = 1.77 \times 10^{-7} M\)
04

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 \(3.32 \times 10^{-5} g\).

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Most popular questions from this chapter

The equilibrium constant for the following reaction is \(1.0 \times 10^{23}\). EDTA is used as a complexing agent in chemical analysis. Solutions of EDTA, usually containing the disodium salt \(\mathrm{Na}_{2} \mathrm{H}_{2} \mathrm{EDTA}\), are used to treat heavy metal poisoning. Calculate \(\left[\mathrm{Cr}^{3+}\right]\) at equilibrium in a solution originally \(0.0010 M\) in \(\mathrm{Cr}^{3+}\) and \(0.050 \mathrm{M}\) in \(\mathrm{H}_{2} \mathrm{EDTA}^{2-}\) and buffered at \(\mathrm{pH}=6.00 .\)

Write equations for the stepwise formation of each of the following complex ions. a. \(\mathrm{CoF}_{6}^{3-}\) b. \(\mathrm{Zn}\left(\mathrm{NH}_{3}\right)_{4}^{2+}\)

What mass of \(\mathrm{ZnS}\left(K_{\text {? }}=2.5 \times 10^{-22}\right.\) ) will dissolve in \(300.0 \mathrm{~mL}\) of \(0.050 \mathrm{M} \mathrm{Zn}\left(\mathrm{NO}_{3}\right)_{2} ?\) Ignore the basic properties of \(\mathrm{S}^{2-}\) .

List some ways one can increase the solubility of a salt in water.

Explain the following phenomenon: You have a test tube with about \(20 \mathrm{~mL}\) silver nitrate solution. Upon adding a few drops of sodium chromate solution, you notice a red solid forming in a relatively clear solution. Upon adding a few drops of a sodium chloride solution to the same test tube, you notice \(a\) white solid and a pale yellow solution. Use the \(K_{s p}\) values in the book to support your explanation, and include the balanced equations.
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