Con base en el valor de la Kp s del Ag2 S, de Ka 1 y Ka 2 del H2 S y Kf=1.1 ×10^5 del AgCl2^-,

step 1 So now we'll work on problem 68 from chapter 17. In this problem, we're told to use the value of

Con base en el valor de la Kp s del Ag2 S, de Ka 1 y Ka 2...

Con base en el valor de la $K_{p s}$ del $\mathrm{Ag}_2 \mathrm{~S}$, de $K_{a 1}$ y $K_{a 2}$ del $\mathrm{H}_2 \mathrm{~S}$ y $K_f=1.1 \times 10^5 \mathrm{del} \mathrm{AgCl}_2^{-}$, calcule la constante de equilibrio de la reacción siguiente: $$ \mathrm{Ag}_2 \mathrm{~S}(\mathrm{~s})+4 \mathrm{Cl}^{-}(a c)+2 \mathrm{H}^{+}(a c) \rightleftharpoons $$ $$ 2 \mathrm{AgCl}_2{ }^{-}(a c)+\mathrm{H}_2 \mathrm{~S}(a c) $$

Con base en el valor de la $K_{p s}$ del $\mathrm{Ag}_2 \mathrm{~S}$, de $K_{a 1}$ y $K_{a 2}$ del $\mathrm{H}_2 \mathrm{~S}$ y $K_f=1.1 \times 10^5 \mathrm{del} \mathrm{AgCl}_2^{-}$, calcule la constante de equilibrio de la reacción siguiente:
$$
\mathrm{Ag}_2 \mathrm{~S}(\mathrm{~s})+4 \mathrm{Cl}^{-}(a c)+2 \mathrm{H}^{+}(a c) \rightleftharpoons
$$
$$
2 \mathrm{AgCl}_2{ }^{-}(a c)+\mathrm{H}_2 \mathrm{~S}(a c)
$$

Key Concepts

Solubility Product (Ksp)

The solubility product constant describes the equilibrium between a solid and its constituent ions in a saturated solution. It quantifies the extent to which a sparingly soluble salt dissolves. In the context of heterogeneous equilibria, the concentration of the solid is taken as a constant, and only the ionic concentrations contribute to the equilibrium expression. It is essential for calculating the dissolution equilibrium of compounds like Ag2S in this reaction.

Acid Dissociation Constant (Ka)

The acid dissociation constant governs the tendency of an acid to donate a proton in solution. For polyprotic acids, such as H2S, multiple dissociation steps occur, each with its own Ka value (Ka1, Ka2, etc.). These constants determine the equilibrium position for the dissociation reactions and are crucial for understanding the acid-base behavior of H2S in the overall reaction.

Formation Constant (Kf)

The formation constant, or stability constant, quantifies the formation of a complex ion from a metal ion and its ligands. In coordination chemistry, a higher Kf indicates that the complex is more stable. In the given problem, the formation of AgCl2? from Ag+ and Cl? is governed by its Kf, which significantly influences the overall equilibrium of the reaction when complexation is involved.

Combination of Equilibrium Constants

The overall equilibrium constant for a net reaction can be obtained by combining the individual equilibrium constants of the sequential reactions that make up the net process. This method involves algebraic multiplication and division of constants, reflecting the stoichiometry and reversibility of the steps involved. Understanding this concept is key to linking the solubility, acid-base, and complex formation equilibria together.

Heterogeneous Equilibrium

Heterogeneous equilibria involve reactants and products in different phases, such as gases, solutions, or solids. In such equilibria, the activity of pure solids and pure liquids is taken as unity, simplifying the equilibrium expressions. This idea is essential when dealing with reactions that include solid salts, like Ag2S, alongside ions in solution.

Transcript

00:01 So now we'll work on problem 68 from chapter 17.

00:08 In this problem, we're told to use the value of ksp for silver sulfide, ka1, ka2 for h2s, and kf equal 1 .1 times 10 of the 5th for silver cl2 minus.

00:25 And we can calculate the equilibrium constant for the following reaction.

00:31 Silver sulfide plus 4 chlorine plus 2 hydrogen makes the silver chloride complex plus hydrogen sulfide.

00:41 So here we need to write down a series of equations that it tells us to use, and we need to be mindful that we're going to be adding them up to equal the equation, the chemical reaction given.

00:53 So we want to write them flipped or multiplied by two or whatever we have to do to make it work out.

01:01 So first we'll work with the ksp.

01:04 For silver sulfide.

01:05 So we have ag2s dissociating to form two silver ions plus a sulfur to minus.

01:18 And the ksp for this reaction is equal to 6 times 10 to the negative 51, very small.

01:29 For the ka1 and ka2 values, we're talking about the dissociation of h2s into h plus and hs minus.

01:43 However, if we look at the bottom, we want to have h2s on the right side.

01:48 So we're going to flip this equation.

01:52 H plus hs minus make h2s.

01:59 So as a reminder, we flipped this.

02:03 So that means that we need to do 1 divided by ka1 because of the flipping.

02:13 And that gives us a value of 1 .05 times 10 to the 7th power.

02:22 Now for the second kaa value, ka2, we want to do the same thing again because now that we have hs minus on the left, we want to have it on the right to cancel out...

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