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Chapter 15: Problem 101

a. Calculate the molar solubility of AgBr in pure water. \(K_{\mathrm{sp}}\) for AgBr is \(5.0 \times 10^{-13}\). b. Calculate the molar solubility of \(\mathrm{AgBr}\) in \(3.0\) \(M\) \(\mathrm{NH}_{3} .\) The overall formation constant for \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}\) is \(1.7 \times 10^{7}\) that is, $$\mathrm{Ag}^{+}(a q)+2 \mathrm{NH}_{3}(a q) \longrightarrow \mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}^{+}(a q) \quad K=1.7 \times 10^{7}$$ c. Compare the calculated solubilities from parts a and b. Explain any differences. d. What mass of AgBr will dissolve in \(250.0 \mathrm{mL}\) of \(3.0 M \mathrm{NH}_{3} ?\) e. What effect does adding \(\mathrm{HNO}_{3}\) have on the solubilities calculated in parts a and b?

Short Answer

Expert verified
The molar solubility of AgBr in pure water is approximately 7.1x10鈦烩伔 M, while its molar solubility in a 3.0 M ammonia solution is about 1.33x10鈦宦光伆 M, which is higher due to the formation of the complex ion Ag(NH鈧)鈧傗伜. The mass of AgBr that will dissolve in 250.0 mL of 3.0 M NH鈧 is approximately 6.25x10鈦宦光伆 g. Adding HNO鈧 will not affect the solubility in pure water but will reduce the solubility of AgBr in ammonia solution as it reacts with NH鈧 and subsequently decreases the concentration of the complex ion.

Step by step solution

01

Write the balanced dissociation reaction for AgBr in water.

AgBr (s) 鈬 Ag鈦 (aq) + Br鈦 (aq)
02

Set up the ICE table

Since the Ksp expression is [Ag鈦篯[Br鈦籡, we will use the ICE table to find the equilibrium concentrations of Ag鈦 and Br鈦. Initial: [Ag鈦篯 = [Br鈦籡 = 0 Change: +s +s Equilibrium: [Ag鈦篯 = [Br鈦籡 = s
03

Calculate the molar solubility, s, using Ksp.

Ksp = [Ag鈦篯[Br鈦籡 = (s)(s) = 5.0x10鈦宦孤 (given) s虏 = 5.0x10鈦宦孤 s (molar solubility) = 鈭(5.0x10鈦宦孤) 鈮 7.1x10鈦烩伔 M b. Calculate the molar solubility of AgBr in 3.0 M NH鈧.
04

Write the balanced reaction for complex formation

Ag鈦 (aq) + 2NH鈧 (aq) 鈬 Ag(NH鈧)鈧傗伜 (aq), K = 1.7x10鈦 (given)
05

Set up the ICE table

Since the K expression for complex formation is [Ag(NH鈧)鈧傗伜]/([Ag鈦篯[NH鈧僝虏), we will use the ICE table to find the equilibrium concentrations. Initial: [Ag鈦篯 - from part a = 7.1x10鈦烩伔 M [NH鈧僝 = 3.0 M (given) [Ag(NH鈧)鈧傗伜] = 0 Change: [-7.1x10鈦烩伔] -2x(+7.1x10鈦烩伔) +7.1x10鈦烩伔 Equilibrium: [0] [2.994] [7.1x10鈦烩伔]
06

Calculate the molar solubility of AgBr using K

K = [Ag(NH鈧)鈧傗伜]/([Ag鈦篯[NH鈧僝虏) = 1.7x10鈦 Solving for [Ag鈦篯, we obtain: [Ag鈦篯 = [Ag(NH鈧)鈧傗伜]/(1.7x10鈦 * (NH鈧)虏) = 7.1x10鈦烩伔/(1.7x10鈦 * 2.994虏) 鈮 1.33x10鈦宦光伆 M c. Compare the calculated solubilities from parts a and b
07

In part a, the molar solubility of AgBr in pure water was 7.1x10鈦烩伔 M, while in part b, the molar solubility in 3.0 M ammonia solution was 1.33x10鈦宦光伆 M. The solubility of AgBr is higher in the ammonia solution due to the formation of a complex ion (Ag(NH鈧)鈧傗伜) which reduces the concentration of free Ag鈦 ions, driving the reaction to the right and dissolving more of the AgBr solid. d. What mass of AgBr will dissolve in 250.0 mL of 3.0 M NH鈧?

Step 1: Calculate the moles of AgBr that will dissolve
08

Moles of AgBr = Molar solubility * Volume = (1.33x10鈦宦光伆 M)(0.250 L) 鈮 3.33x10鈦宦孤 mol

Step 2: Calculate the mass of AgBr using its molar mass
09

Mass of AgBr = moles * molar mass = (3.33x10鈦宦孤 mol) * (187.77 g/mol; molar mass of AgBr) 鈮 6.25x10鈦宦光伆 g e. What effect does adding HNO鈧 have on the solubilities calculated in parts a and b?

Adding HNO鈧 (a strong acid) will increase the concentration of H鈦 ions in the solution. In part a (pure water), adding HNO鈧 will not affect the AgBr solubility significantly as there is no equilibrium involving H鈦 ions in the dissociation of AgBr. In part b (ammonia solution), however, adding HNO鈧 will react with NH鈧 and form NH鈧勨伜 ions, thus reducing the concentration of NH鈧 available to form the complex ion Ag(NH鈧)鈧傗伜. This will shift the equilibrium of complex formation to the left, increasing the concentration of Ag鈦 ions and, consequently, reducing the solubility of AgBr.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ksp (Solubility Product Constant)
Understanding the Solubility Product Constant, or Ksp, is key to predicting the solubility of ionic compounds in water. Ksp is an equilibrium constant specific to the dissolution of solids. It represents the level at which a solute's ions are saturated in solution, beyond which any additional solute will precipitate.

Taking the example of AgBr dissolving in water, where AgBr (s) 鈬 Ag鈦 (aq) + Br鈦 (aq), we see that the Ksp equation becomes Ksp = [Ag鈦篯[Br鈦籡. Here, the concentrations are those at equilibrium, where the rates of dissolution and precipitation are equal.

In practice, if you know the Ksp and the stoichiometry of the dissolution, you can calculate the molar solubility of the compound. For AgBr, with a given Ksp of 5.0 x 10鈦宦孤, the molar solubility is the square root of the Ksp, assuming the dissolution produces equal molar amounts of Ag鈦 and Br鈦. The lower the Ksp value, the lower the solubility of the compound in water.
Complex Ion Formation
Moving beyond simple solubility, complex ion formation can significantly alter solubility patterns. Complex ions are species formed from a metal ion and one or more ligands, which are molecules or ions that can donate an electron pair to the metal ion.When a metal ion in solution, such as Ag鈦, encounters ligands like NH鈧, they can form a complex ion, in this case Ag(NH鈧)鈧傗伜. The formation of such complex ions is highly relevant because it can greatly increase the solubility of an original compound. A new equilibrium constant, called the formation constant (Kf), is used to represent the stability of the complex ion.In the presence of excess NH鈧, the reaction Ag鈦 (aq) + 2NH鈧 (aq) 鈬 Ag(NH鈧)鈧傗伜 (aq) has its own equilibrium constant, Kf, which is significantly larger than Ksp. The effect is a larger dissolution of AgBr than would occur without NH鈧, due to the formation of the stable Ag(NH鈧)鈧傗伜 complex.
ICE Table Method
To quantify the dynamic between solute and solvent, or complex ion formation, chemists use an ICE table method鈥攕tanding for Initial, Change, and Equilibrium. It鈥檚 a systematic way to organize the concentrations of reactants and products in an equilibrium reaction. For the solubility of AgBr in water, we begin with the concentrations of Ag鈦 and Br鈦 as zero (Initial). When AgBr starts dissolving (Change), both Ag鈦 and Br鈦 increase by s, where s is the solubility of AgBr. At Equilibrium, the concentrations are equal to s. Then, applying the Ksp expression allows us to solve for s.When considering complex ion formation, the ICE table adjusts to include the complex ion and the ligand. For AgBr in NH鈧, the initial concentration of Ag鈦 is the solubility from the water-only scenario, and NH鈧 has a known concentration. As NH鈧 reacts with Ag鈦 (Change), we track the concentrations until reaching Equilibrium. The calculations here are more intricate, as they require accounting for the Kf of the complex ion. The ICE table method neatly handles these, allowing chemists to solve for the unknowns in multi-step equilibria.

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Most popular questions from this chapter

Will a precipitate form when \(100.0 \mathrm{mL}\) of \(4.0 \times 10^{-4} \mathrm{M}\) \(\mathrm{Mg}\left(\mathrm{NO}_{3}\right)_{2}\) is added to \(100.0 \mathrm{mL}\) of \(2.0 \times 10^{-4}\) \(M\) \(\mathrm{NaOH} ?\)

Consider \(1.0 \mathrm{L}\) of an aqueous solution that contains \(0.10\) \(M\) sulfuric acid to which 0.30 mole of barium nitrate is added. Assuming no change in volume of the solution, determine the \(\mathrm{pH},\) the concentration of barium ions in the final solution, and the mass of solid formed.

Calculate the concentration of \(\mathrm{Pb}^{2+}\) in each of the following. a. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}, K_{\mathrm{sp}}=1.2 \times 10^{-15}\) b. a saturated solution of \(\mathrm{Pb}(\mathrm{OH})_{2}\) buffered at \(\mathrm{pH}=13.00\) c. Ethylenediaminetetraacetate (EDTA \(^{4-}\) ) is used as a complexing agent in chemical analysis and has the following structure: Solutions of EDTA \(^{4-}\) are used to treat heavy metal poisoning by removing the heavy metal in the form of a soluble complex ion. The reaction of EDTA \(^{4-}\) with \(\mathrm{Pb}^{2+}\) is $$\begin{aligned} \mathrm{Pb}^{2+}(a q)+\mathrm{EDTA}^{4-}(a q) \rightleftharpoons \mathrm{PbEDTA}^{2-}(a q) & \\ K &=1.1 \times 10^{18} \end{aligned}$$ Consider a solution with 0.010 mole of \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) added to \(1.0 \mathrm{L}\) of an aqueous solution buffered at \(\mathrm{pH}=13.00\) and containing 0.050 \(M\) \(\mathrm{Na}_{4} \mathrm{EDTA}\). Does \(\mathrm{Pb}(\mathrm{OH})_{2}\) precipitate from this solution?

The solubility of \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\) in a \(7.2 \times 10^{-2}-M \mathrm{KIO}_{3}\) solution is \(6.0 \times 10^{-9} \mathrm{mol} / \mathrm{L} .\) Calculate the \(K_{\mathrm{sp}}\) value for \(\mathrm{Pb}\left(\mathrm{IO}_{3}\right)_{2}(s)\).

\(\mathrm{Ag}_{2} \mathrm{S}(s)\) has a larger molar solubility than CuS even though \(\mathrm{Ag}_{2} \mathrm{S}\) has the smaller \(K_{\mathrm{sp}}\) value. Explain how this is possible.
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