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Chapter 16: Problem 51

For the reaction $$ \mathrm{CdC}_{2} \mathrm{O}_{4}(s)+4 \mathrm{NH}_{3}(a q) \rightleftharpoons \mathrm{Cd}\left(\mathrm{NH}_{3}\right)_{4}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) $$ (a) calculate \(K .\left(K_{s p}\right.\) for \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is \(\left.1.5 \times 10^{-8} .\right)\) (b) calculate \(\left[\mathrm{NH}_{3}\right]\) at equilibrium when \(2.00 \mathrm{~g}\) of \(\mathrm{CdC}_{2} \mathrm{O}_{4}\) is dissolved in \(1.00 \mathrm{~L}\) of solution.

Short Answer

Expert verified
Answer: The concentration of NH3 at equilibrium is 0.141 mol/L.

Step by step solution

01

Understanding the relationship between K and Ksp

For the given reaction, we have a solid substance (CdC2O4) dissolving in water and forming a complex ion (Cd(NH3)4^(2+)) and an anion (C2O4^(2-)). We are provided with the solubility product constant, Ksp = 1.5 x 10^(-8), for CdC2O4. From this information, we can determine the equilibrium constant (K) of the given reaction. Remember that: $$ K = \frac{[Cd(NH_3)_4^{2+}][C_2O_4^{2-}]}{[CdC_2O_4][NH_3]^4} = K_{sp} $$ Since [CdC2O4] is a solid, its concentration can be considered constant, and it's not included in the equation. Therefore, K = Ksp.
02

Finding K for the reaction

Since Ksp = 1.5 x 10^(-8), we have: $$ K = 1.5 \times 10^{-8} $$ Step 2: Calculate the concentration of NH3 at equilibrium
03

Convert the mass of CdC2O4 to moles

In order to find the concentration of NH3 at equilibrium, we first need to convert the mass of CdC2O4 to moles. Given that we have 2.00 g of CdC2O4, the molar mass of CdC2O4 is 183.5 g/mol. Therefore: $$ \text{moles of CdC}_2\text{O}_4 = \frac{2.00 \text{ g}}{183.5 \text{ g/mol}} = 0.0109 \text{ mol} $$
04

Using the ICE table to determine the equilibrium concentration of NH3

Next, we will use an ICE (Initial, Change, Equilibrium) table to find the NH3 concentration at equilibrium. Let 'x' be the amount of CdC2O4 that dissociates. Initial concentrations: $$[Cd(NH_3)_4^{2+}]=0$$ $$[C_2O_4^{2-}]=0$$ $$[NH_3]= y$$ Changes: $$[Cd(NH_3)_4^{2+}]=+x$$ $$[C_2O_4^{2-}]=+x$$ $$[NH_3] = -4x$$ Equilibrium concentrations: $$[Cd(NH_3)_4^{2+}]=x$$ $$[C_2O_4^{2-}]=x$$ $$[NH_3] = y - 4x$$ Now we can use the K expression from Step 1 along with the information from the ICE table: $$ K = \frac{[Cd(NH_3)_4^{2+}][C_2O_4^{2-}]}{[NH_3]^4} = \frac{x \cdot x}{(y - 4x)^4} $$ Since x is very small compared to y, we can approximate and assume that 4x<<y so (y-4x) 鈮� y. Substitute K = 1.5 x 10^(-8) and solve for x: $$ 1.5 \times 10^{-8} = \frac{x^2}{y^4} $$ We know that the number of moles of CdC2O4 that dissolved is equal to x, and since 1.00L of solution: $$ \frac{0.0109}{1.00} = x = 0.0109 \text{ mol/L} $$ Now substitute x back into our simplified equilibrium and solve for the initial concentration of NH3 (y) $$ 1.5 \times 10^{-8} = \frac{(0.0109)^2}{y^4} $$ $$ y^4 = \frac{(0.0109)^2}{1.5 \times 10^{-8}} $$ $$ y = 0.141 \text{ mol/L} $$ So, the concentration of NH3 at equilibrium is 0.141 mol/L.

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Key Concepts

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

Solubility Product Constant (Ksp)
The Solubility Product Constant, often abbreviated as Ksp, is a critical concept when dealing with the solubility of compounds. It is used to describe the equilibrium between a solid and its respective ions in a saturated solution. It represents the extent to which a solid can dissolve in water.
When a slightly soluble ionic compound dissolves, it dissociates into its constituent ions. For example, a generic salt AB, will dissociate into A鈦� and B鈦� ions.
The expression for Ksp is formulated based on the concentrations of these ions at equilibrium:
  • For a compound like AB, the Ksp expression would be: \[K_{sp} = [A^+][B^-]\]
  • In the case of a solid dissociating to produce multiple ions, the Ksp expression will include powers corresponding to the stoichiometry of the dissolution reaction.

For the given reaction in our exercise, although the solid cadmium oxalate, CdC鈧侽鈧�, doesn't directly appear in the equilibrium equation, its Ksp is vital for finding the equilibrium constant by balancing the ions in solution.
Molar Mass Calculation
Molar mass is a fundamental parameter in chemistry that connects the mass of a compound to the amount of substance. It's the mass of one mole of a given substance and is usually expressed in grams per mole (g/mol).
When you know the molar mass, you can easily convert between the mass of a substance and the moles, which is often a necessary step in chemical calculations.
For instance, if you have a compound such as cadmium oxalate (CdC鈧侽鈧�), to find its molar mass, you sum the mass of each constituent element based on its atomic mass and the number of atoms:
  • Cadmium (Cd) has an atomic mass of 112.4 g/mol.
  • Carbon (C) has an atomic mass of 12.01 g/mol and there are 2 carbon atoms in oxalate.
  • Oxygen (O) has an atomic mass of 16.00 g/mol and there are 4 oxygen atoms.

This gives you: \[112.4 + (2 \times 12.01) + (4 \times 16.00) = 183.42 \, \text{g/mol} \]
Converting the given substance mass to moles is straightforward:
  • If you have 2.00 g of CdC鈧侽鈧�, you calculate the moles as follows:
  • \[\text{moles} = \frac{2.00 \, \text{g}}{183.42 \, \text{g/mol}} \]
  • This results in approximately 0.0109 mol of CdC鈧侽鈧�.
ICE Table Calculation
The ICE table is a helpful tool for organizing data and calculating concentrations of reactants and products at equilibrium. It stands for Initial, Change, Equilibrium, reflecting the three stages of concentration during a reaction.
Using an ICE table:
  • Identify the initial concentrations of all species present before the reaction reaches equilibrium.
  • Define the changes that occur as the reaction proceeds to equilibrium. This is commonly represented by x, which is the change in concentration.
  • Finally, calculate the equilibrium concentrations by combining the initial concentrations and the changes.

For the reaction in our exercise, when cadmium oxalate dissolves in the presence of ammonia, we can use the ICE table setup:
  • Initial condition: [Cd(NH鈧�)鈧劼测伜] = 0, [C鈧侽鈧劼测伝] = 0, and ammonia starts at an unknown concentration y.
  • Change: As x amount of cadmium oxalate dissolves, it generates x moles of Cd(NH鈧�)鈧劼测伜 and C鈧侽鈧劼测伝 each, while ammonia decreases by 4x.
  • Equilibrium: This gives us x for the ions and y - 4x for ammonia.

By substituting these into the equilibrium expression of K:\[K = \frac{(x)\times(x)}{(y - 4x)^4}\]
considering x is small, simplifies the model to solve for the initial concentration, yielding precise values for equilibrium scenarios.

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Most popular questions from this chapter

What is the solubility of \(\mathrm{CaF}_{2}\) in a buffer solution containing \(0.30 \mathrm{M}\) \(\mathrm{HCHO}_{2}\) and \(0.20 \mathrm{M} \mathrm{NaCHO}_{2}\) ? Hint: Consider the equation $$ \mathrm{CaF}_{2}(s)+2 \mathrm{H}^{+}(a q) \longrightarrow \mathrm{Ca}^{2+}(a q)+2 \mathrm{HF}(a q) $$ and solve the equilibrium problem.

Predict what effect each of the following has on the position of the equilibrium $$ \mathrm{PbCl}_{2}(s) \rightleftharpoons \mathrm{Pb}^{2+}(a q)+2 \mathrm{Cl}^{-}(a q) \quad \Delta H=23.4 \mathrm{~kJ} $$ (a) addition of \(1 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) solution (b) increase in temperature (c) addition of \(\mathrm{Ag}^{+}\), forming \(\mathrm{AgCl}\) (d) addition of \(1 M\) hydrochloric acid

Consider the insoluble salts \(\mathrm{JQ} \mathrm{K}_{2} \mathrm{R}, \mathrm{L}_{2} \mathrm{~S}_{3}, \mathrm{MT}_{2}\), and \(\mathrm{NU}_{3}\). They are formed from the metal ions \(\mathrm{J}^{+}, \mathrm{K}^{+}, \mathrm{L}^{3+}, \mathrm{M}^{2+}\), and \(\mathrm{N}^{3+}\) and the nonmetal ions \(\mathrm{Q}^{-}, \mathrm{R}^{2-}, \mathrm{S}^{2-}, \mathrm{T}^{-}\), and \(\mathrm{U}^{-}\). All the salts have the same \(K_{\text {sp }}, 1 \times 10^{-10}\), at \(25^{\circ} \mathrm{C}\). (a) Which salt has the highest molar solubility? (b) Does the salt with the highest molar solubility have the highest solubility in g salt/100 g water? (c) Can the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water be determined from the information given? If yes, calculate the solubility of each salt in \(\mathrm{g} / 100 \mathrm{~g}\) water. If no, why not?

Consider the following solubility data for calcium oxalate \(\left(\mathrm{Ca} \mathrm{C}_{2} \mathrm{O}_{4}\right)\) : $$ \begin{aligned} &K_{s p} \text { at } 25^{\circ} \mathrm{C}=4 \times 10^{-9} \\ &K_{s p} \text { at } 95^{\circ} \mathrm{C}=1 \times 10^{-8} \end{aligned} $$ Five hundred mL of a saturated solution is prepared at \(95^{\circ} \mathrm{C}\). How many milligrams of \(\mathrm{CaC}_{2} \mathrm{O}_{4}\) will precipitate when the solution is cooled to \(25^{\circ} \mathrm{C}\) ? (Assume that supersaturation does not take place.)

Calcium ions in blood trigger clotting. To prevent that in donated blood, sodium oxalate, \(\mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), is added to remove calcium ions according to the following equation. $$ \mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q)+\mathrm{Ca}^{2+}(a q) \longrightarrow \mathrm{CaC}_{2} \mathrm{O}_{4}(s) $$ Blood contains about \(0.10 \mathrm{mg} \mathrm{Ca}^{2+} / \mathrm{mL}\). If a 250.0-mL sample of donated blood is treated with an equal volume of \(0.160 \mathrm{M} \mathrm{Na}_{2} \mathrm{C}_{2} \mathrm{O}_{4}\), estimate \(\left[\mathrm{Ca}^{2+}\right]\) after precipitation. \(\left(K_{s p} \mathrm{CaC}_{2} \mathrm{O}_{4}=4 \times 10^{-9}\right)\)
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