## Chemistry: The Molecular Science (5th Edition)

Published by Cengage Learning

# Chapter 15 - Additional Aqueous Equilibria - Problem Solving Practice 15.13 - Page 687: a

#### Answer

$K_{net} = 3.6 \times 10^{3}$ Since the constant value is greater than 1, the reaction is product favored.

#### Work Step by Step

1. Write the $K_{sp}$ chemical reaction: $AgCl(s) \lt -- \gt Ag^+(aq) + Cl^-(aq)$ _____ $K_{sp} = 1.8 \times 10^{-10}$ 2. Write the chemical reaction that produces $[Ag(S_2O_3)_2]^{3-}$: $Ag^+(aq) + 2S_2O{_3}^{2-}(aq) \lt -- \gt [Ag(S_2O_3)_2]^{3-}(aq)$ $K_f = 2.0 \times 10^{13}$ 3. Now, combine these equations: $AgCl(s) + Ag^+(aq) + 2S_2O{_3}^{2-}(aq) \lt -- \gt Ag^+(aq) + Cl^-(aq)+ [Ag(S_2O_3)_2]^{3-}(aq)$ - When you do this, you will have to multiply the equilibrium constants: $K_{net} = 1.8 \times 10^{-10} \times 2.0 \times 10^{13} = 3.6 \times 10^{3}$ Since the constant value is greater than 1, the reaction is product favored:

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