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Chapter 18: Problem 40

At \(20^{\circ} \mathrm{C},\) a saturated aqueous solution of silver acetate, \(\mathrm{AgCH}_{3} \mathrm{CO}_{2},\) contains \(1.0 \mathrm{g}\) of the silver compound dissolved in \(100.0 \mathrm{mL}\) of solution. Calculate \(K_{\mathrm{sp}}\) for silver acetate. $$\mathrm{AgCH}_{3} \mathrm{CO}_{2}(\mathrm{s}) \rightleftarrows \mathrm{Ag}^{+}(\mathrm{aq})+\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})$$

Short Answer

Expert verified
The solubility product constant \(K_{\mathrm{sp}}\) for silver acetate is approximately 0.00358.

Step by step solution

01

Calculate Moles of Silver Acetate

To find the number of moles, we first need to know the molar mass of silver acetate (\(\text{AgCH}_3\text{CO}_2\)). The molar mass is calculated as follows:- Ag: 107.87 g/mol- C (2 atoms): 12.01 g/mol \(\times 2 = 24.02\)- H (3 atoms): 1.01 g/mol \(\times 3 = 3.03\)- O (2 atoms): 16.00 g/mol \(\times 2 = 32.00\)Adding these gives:\[\mathrm{Molar\ Mass\ of\ AgCH}_3 \mathrm{CO}_2 = 107.87 + 24.02 + 3.03 + 32.00 = 166.92 \mathrm{\ g/mol}\]Now, calculate the moles of silver acetate using the given mass (1.0 g):\[\text{Moles of AgCH}_3\text{CO}_2 = \frac{1.0\ \text{g}}{166.92\ \text{g/mol}} \approx 0.00599\ \text{mol}\]
02

Find Molarity of Ions

The solution volume is 100 mL, which is 0.1 L. The molarity (concentration) of silver ions (\(\mathrm{Ag}^+\)) and acetate ions (\(\mathrm{CH}_3\mathrm{CO}_2^-\)) is the same as that of the dissolved silver acetate:\[\text{Molarity of ions} = \frac{0.00599\ \text{mol}}{0.1\ \text{L}} = 0.0599\ \text{M}\]
03

Write Expression for Ksp

The solubility product constant \(K_{\mathrm{sp}}\) is expressed in terms of the ion concentrations at equilibrium. For silver acetate dissociation:\[\mathrm{AgCH}_3\mathrm{CO}_2(s) \rightleftarrows \mathrm{Ag}^+(aq) + \mathrm{CH}_3\mathrm{CO}_2^-(aq)\]\(K_{\mathrm{sp}}\) is given by:\[K_{\mathrm{sp}} = [\mathrm{Ag}^+][\mathrm{CH}_3\mathrm{CO}_2^-]\]\(\text{Where } [\mathrm{Ag}^+] = [\mathrm{CH}_3\mathrm{CO}_2^-] = 0.0599\ \text{M}\).
04

Calculate Ksp

Substitute the concentrations into the \(K_{\mathrm{sp}}\) expression:\[K_{\mathrm{sp}} = (0.0599)(0.0599) = (0.0599)^2\]Calculating gives:\[K_{\mathrm{sp}} \approx 0.00358\]

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Key Concepts

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

Understanding Molarity
Molarity is a crucial concept in chemistry that refers to the concentration of a solute in a solution. It's important because it helps us understand how much of a substance is present in a given volume. The formula for molarity is:
  • M = \( \frac{\text{moles of solute}}{\text{liters of solution}} \)
With the silver acetate solution in our exercise, we found 0.00599 moles were dissolved in 0.1 liters of water. Thus, the molarity or concentration of the silver acetate is 0.0599 M.
This means for every liter of solution, there are 0.0599 moles of dissolved solute.
Understanding molarity is key to predicting how substances will react, especially when calculating specific equilibrium constants.
Equilibrium Expression Essentials
An equilibrium expression describes the relationship between the concentrations of reactants and products at equilibrium for a reversible chemical reaction.
In our case, the dissociation of silver acetate can be represented as follows:
  • \( \text{AgCH}_3\text{CO}_2(s) \rightleftharpoons \text{Ag}^+(aq) + \text{CH}_3\text{CO}_2^-(aq) \)
The equilibrium expression to find the solubility product constant \( K_{\text{sp}} \) is:
  • \( K_{\text{sp}} = [\text{Ag}^+][\text{CH}_3\text{CO}_2^-] \)
At equilibrium, both ion concentrations are equal, leading to \( K_{\text{sp}} = (0.0599)^2 \). This highlights the importance of equilibrium in understanding how concentrations balance out in saturated solutions.
Calculating Molar Mass
To understand molarity and ultimately balance chemical equations, molar mass is essential. It's the mass of one mole of a given substance and is calculated by summing the atomic masses of all the atoms in a formula.
For silver acetate, the molar mass calculation involves:
  • Silver (Ag): 107.87 g/mol
  • Carbon (C): 12.01 g/mol \( \times 2 = 24.02 \)
  • Hydrogen (H): 1.01 g/mol \( \times 3 = 3.03 \)
  • Oxygen (O): 16.00 g/mol \( \times 2 = 32.00 \)
Adding these values gives us a total molar mass of 166.92 g/mol. This is crucial for determining the exact number of moles in a given mass, thus allowing us to compute concentration accurately.

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Most popular questions from this chapter

Calculate the hydronium ion concentration and the pH of the solution that results when \(20.0 \mathrm{mL}\) of \(0.15 \mathrm{M}\) acetic acid, \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H},\) is mixed with \(5.0 \mathrm{mL}\) of \(0.17 \mathrm{M} \mathrm{NaOH}\).

When \(1.55 \mathrm{g}\) of solid thallium (I) bromide is added to \(1.00 \mathrm{L}\) of water, the salt dissolves to a small extent. $$\operatorname{TIBr}(\mathrm{s}) \rightleftarrows \mathrm{TI}^{+}(\mathrm{aq})+\mathrm{Br}^{-}(\mathrm{aq})$$ The thallium(I) and bromide ions in equilibrium with TIBr each have a concentration of \(1.9 \times 10^{-3} \mathrm{M} .\) What is the value of \(K_{\mathrm{sp}}\) for TIBr?

You dissolve \(0.425 \mathrm{g}\) of \(\mathrm{NaOH}\) in \(2.00 \mathrm{L}\) of a buffer solution that has \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]=\left[\mathrm{HPO}_{4}^{2-}\right]=0.132 \mathrm{M} .\) What is the pH of the solution before adding NaOH? After adding NaOH?

Lactic acid (CH \(_{3} \mathrm{CHOHCO}_{2} \mathrm{H}\) ) is found in sour milk, in sauerkraut, and in muscles after activity (see page 479 ). \(\left(K_{\mathrm{a}} \text { for lactic acid }=1.4 \times 10^{-4} .\right)\) (a) If 2.75 g of \(\mathrm{NaCH}_{3} \mathrm{CHOHCO}_{2},\) sodium lactate, is added to \(5.00 \times 10^{2} \mathrm{mL}\) of \(0.100 \mathrm{M}\) lactic acid, what is the \(\mathrm{pH}\) of the resulting buffer solution? (b) Is the final pH lower or higher than the pH of the lactic acid solution?

Calculate the pH of a solution that has an acetic acid concentration of \(0.050 \mathrm{M}\) and a sodium acetate concentration of \(0.075 \mathrm{M}.\)
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