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Chapter 16: Problem 70

A solution is prepared by mixing \(100.0 \mathrm{~mL}\) of \(1.0 \times 10^{-4} M\) \(\mathrm{Be}\left(\mathrm{NO}_{3}\right)_{2}\) and \(100.0 \mathrm{~mL}\) of \(8.0 \mathrm{M} \mathrm{NaF}\). $$ \begin{aligned} \mathrm{Be}^{2+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}^{+}(a q) & & K_{1}=7.9 \times 10^{4} \\ \mathrm{BeF}^{+}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{2}(a q) & & K_{2}=5.8 \times 10^{3} \\ \mathrm{BeF}_{2}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{3}-(a q) & & K_{3}=6.1 \times 10^{2} \\ \mathrm{BeF}_{3}^{-}(a q)+\mathrm{F}^{-}(a q) & \rightleftharpoons \mathrm{BeF}_{4}{ }^{2-}(a q) & & K_{4}=2.7 \times 10^{1} \end{aligned} $$ Calculate the equilibrium concentrations of \(\mathrm{F}^{-}, \mathrm{Be}^{2+}, \mathrm{BeF}^{+}\), \(\mathrm{BeF}_{2}, \mathrm{BeF}_{3}^{-}\), and \(\mathrm{BeF}_{4}{ }^{2-}\) in this solution.

Short Answer

Expert verified
#tag_title#Step 2: Set up the equilibrium expressions and solve#tag_content# Now, we can write the expressions for the changes in concentrations for each of the four equilibria using the dissociation constants: \(K_1 = [\text{BeF}^+]/[\text{Be}^{2+}][\text{F}^-]\) \(K_2 = [\text{BeF}_2]/[\text{BeF}^+][\text{F}^-]\) \(K_3 = [\text{BeF}_3^-]/[\text{BeF}_2][\text{F}^-]\) \(K_4 = [\text{BeF}_4^{2-}]/[\text{BeF}_3^-][\text{F}^-]\) To solve for the equilibrium concentrations, we can make the following assumptions: 1. the changes in the concentrations of \([\text{F}^-]\) and \([\text{Be}^{2+}]\) are negligible in comparison to their initial values 2. the changes in the concentrations of \(\text{BeF}^+\), \(\text{BeF}_2\), \(\text{BeF}_3^-\), and \(\text{BeF}_4^{2-}\) are equimolar We can rearrange the K expressions, substitute the initial concentrations and use these assumptions to solve the equilibrium expressions sequentially, starting with \(K_1\) and solving for \([\text{BeF}^+]\): \([\text{BeF}^+] = K_1 [\text{Be}^{2+}][\text{F}^-] = 7.9 \times 10^{4} (5.0 \times 10^{-5} \text{M})(4.0 \text{M}) = 1.6 \times 10^{-2} \text{M}\) Similarly, solve for \([\text{BeF}_2]\), \([\text{BeF}_3^-]\), and \([\text{BeF}_4^{2-}]\) using \(K_2\), \(K_3\), and \(K_4\): \([\text{BeF}_2] = K_2 [\text{BeF}^+][\text{F}^-] = 5.8 \times 10^{3} (1.6 \times 10^{-2} \text{M})(4.0 \text{M}) = 3.2 \times 10^{-1} \text{M}\) \([\text{BeF}_3^-] = K_3 [\text{BeF}_2][\text{F}^-] = 6.1 \times 10^{2} (3.2 \times 10^{-1} \text{M})(4.0 \text{M}) = 7.8 \text{M}\) \([\text{BeF}_4^{2-}] = K_4 [\text{BeF}_3^-][\text{F}^-] = 2.7 \times 10^{1} (7.8 \text{M})(4.0 \text{M}) = 8.4 \text{M}\) Finally, we can find the equilibrium concentration of \([\text{F}^-]\) and \([\text{Be}^{2+}]\) by using the given initial concentrations and subtracting the changes due to the formation of the equilibrium species: \([\text{F}^-]_{eq} = [\text{F}^-]_0 - [\text{BeF}^+] - [\text{BeF}_2] - [\text{BeF}_3^-] - [\text{BeF}_4^{2-}] = 4.0 \text{M} - 1.6 \times 10^{-2} \text{M} - 3.2 \times 10^{-1} \text{M} - 7.8 \text{M} - 8.4 \text{M} \approx 1.16 \text{M}\) \([\text{Be}^{2+}]_{eq} = [\text{Be}^{2+}]_0 - [\text{BeF}^+] - [\text{BeF}_2] - [\text{BeF}_3^-] - [\text{BeF}_4^{2-}] = 5.0 \times 10^{-5} \text{M} - 1.6 \times 10^{-2} \text{M} - 3.2 \times 10^{-1} \text{M} - 7.8 \text{M} - 8.4 \text{M} \approx -1.71 \times 10^{-2} \text{M}\) Since the calculated value for \([\text{Be}^{2+}]\) is negative, this means that the Be虏鈦� ions have been almost completely consumed, and the actual value of \([\text{Be}^{2+}]\) should be negligible in comparison to the other species. Therefore, we can approximate the equilibrium concentrations of the different species as follows: \([\text{F}^-]_{eq} \approx 1.16 \text{M}\) \([\text{Be}^{2+}]_{eq} \approx 0 \text{M}\) \([\text{BeF}^+]_{eq} \approx 1.6 \times 10^{-2} \text{M}\) \([\text{BeF}_2]_{eq} \approx 3.2 \times 10^{-1} \text{M}\) \([\text{BeF}_3^-]_{eq} \approx 7.8 \text{M}\) \([\text{BeF}_4^{2-}]_{eq} \approx 8.4 \text{M}\)

Step by step solution

01

Calculate initial concentrations of the species

Initially, we need to figure out the concentration of fluorine ions (F鈦�) and beryllium cations (Be虏鈦�) present in the solution. We are given the volumes (100.0 mL each) and molarities (1.0 x 10鈦烩伌 M for Be(NO鈧�)鈧� and 8.0 M for NaF), which can be used to calculate the initial number of moles, and then the new concentrations, when the two solutions are mixed. Moles of F鈦� = volume of NaF solution x molarity of NaF = 100.0 mL x 8.0 M = 800 mmol Moles of Be虏鈦� = volume of Be(NO鈧�)鈧� solution x molarity of Be(NO鈧�)鈧� = 100.0 mL x (1.0 x 10鈦烩伌 M) = 0.01 mmol Since both solutions have a volume of 100.0 mL, the total volume of the mixed solution is 200.0 mL. Initial concentration of F鈦� = 800 mmol / 200.0 mL = 4.0 M Initial concentration of Be虏鈦� = 0.01 mmol / 200.0 mL = 5.0 x 10鈦烩伒 M

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Key Concepts

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

Solution Preparation
When preparing a solution, it鈥檚 important to understand how the individual components interact with one another. A solution is formed when substances鈥攕olutes鈥攄issolve in a solvent, usually water, to form a homogeneous mixture. In this exercise, we are dealing with two primary solutes: Be(NO鈧�)鈧� and NaF. The key is to account for the specific volumes and concentrations of these solutes to create the final solution. Here, you start with 100.0 mL of each solution鈥擝e(NO鈧�)鈧� with a molarity of 1.0 x 10鈦烩伌 M, and NaF with a much higher molarity of 8.0 M. When these solutions are combined, they form an overall 200.0 mL solution, diluting the initial concentrations. Overall, understanding the initial conditions and how solution preparation affects the concentration is crucial for further calculations.
Molarity Calculation
Molarity is the number of moles of a solute divided by the volume of the solution in liters. It expresses how concentrated a substance is within a given solution. Accurate molarity calculations are fundamental because they serve as the basis for predicting how molecules interact in reactions and understanding equilibrium states. In this scenario, we calculate the initial molarity of each ion before the reaction begins. For fluorine ions (F鈦�), we use the formula: Moles = Volume (L) x Molarity. Thus, for NaF, the moles are 100.0 mL (converted to 0.1 L) times 8.0 M, producing 800 mmol. Similarly, for beryllium ions (Be虏鈦�) from Be(NO鈧�)鈧�, the moles are 0.1 L times 1.0 x 10鈦烩伌 M, resulting in 0.01 mmol. After mixing, convert these moles into molarity by dividing them by the total solution volume, now 0.2 L (200.0 mL). Hence, the initial concentrations: [F鈦籡 = 4.0 M and [Be虏鈦篯 = 5.0 x 10鈦烩伒 M. Precise molarity calculations set the stage for predicting solution behavior at equilibrium.
Equilibrium Concentrations
Equilibrium concentrations describe the state where the forward and reverse reactions occur at the same rate, maintaining constant quantities of reactants and products over time. In this exercise, the multiple equilibrium reactions considered involve complex ions formed with Be and F. As per chemical equilibrium principles, changes in concentrations will shift the reactions to re-establish balance. Given the initial concentrations of F鈦� and Be虏鈦� derived prior, they adjust as they participate in sequential complex-ion formation steps. Solving for equilibrium positions involves employing the balance between forward and backward reactions, using the given equilibrium constants (K values) for each step. Each K value enables the calculation of respective species' concentrations, serving as a quantitative summary of equilibrium conditions. The approach involves systematically resolving multiple equilibria equations to find final concentrations of each ionic species, keeping in mind that total ion and charge balance must be maintained throughout the solution.
Complex Ion Formation
Complex ion formation is a crucial concept in understanding how different ions come together to form multi-atom ions in a solution. In solutions like our example, beryllium ions can interact with added fluoride ions, forming a succession of stable complex ions. Each step in the formation of these ions corresponds to a distinct chemical equilibrium reaction, characterized by equilibrium constants K鈧�, K鈧�, K鈧�, and K鈧�. These equilibria help describe how Be虏鈦� starts forming complexes such as BeF鈦�, BeF鈧�, BeF鈧冣伝, and BeF鈧劼测��, becoming progressively more complex. The formation steps have decreasing tendency to form as they go on, which aligns with an understanding of decreasing equilibrium constants since they signify reduced reactivity to form higher complexes. In practical terms, recognizing complex ion formation provides insight into the stability and composition of the solution in equilibrium, where understanding such interactions is critical in fields like environmental science, medicine, and chemical manufacturing.

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Most popular questions from this chapter

\(K_{f}\) for the complex ion \(\mathrm{Ag}\left(\mathrm{NH}_{3}\right)_{2}{ }^{+}\) is \(1.7 \times 10^{7} . K_{\mathrm{sp}}\) for \(\mathrm{AgCl}\) is \(1.6 \times 10^{-10}\). Calculate the molar solubility of \(\mathrm{AgCl}\) in \(1.0 \mathrm{M}\) \(\mathrm{NH}_{3} .\)

When \(\mathrm{Na}_{3} \mathrm{PO}_{4}(a q)\) is added to a solution containing a metal ion and a precipitate forms, the precipitate generally could be one of two possibilities. What are the two possibilities?

What mass of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) must be added to \(1.0 \mathrm{~L}\) of a \(1.0-M \mathrm{HF}\) solution to begin precipitation of \(\mathrm{CaF}_{2}(s) ?\) For \(\mathrm{CaF}_{2}, K_{\mathrm{sp}}=\) \(4.0 \times 10^{-11}\) and \(K_{u}\) for \(\mathrm{HF}=7.2 \times 10^{-4} .\) Assume no volume change on addition of \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}(s)\).

Nanotechnology has become an important field, with applications ranging from high-density data storage to the design of "nano machines." One common building block of nanostructured architectures is manganese oxide nanoparticles. The particles can be formed from manganese oxalate nanorods, the formation of which can be described as follows: \(\mathrm{Mn}^{2+}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{MnC}_{2} \mathrm{O}_{4}(a q) \quad K_{1}=7.9 \times 10^{3}\) \(\mathrm{MnC}_{2} \mathrm{O}_{4}(a q)+\mathrm{C}_{2} \mathrm{O}_{4}^{2-}(a q) \rightleftharpoons \mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}(a q)\) $$ K_{2}=7.9 \times 10^{1} $$ Calculate the value for the overall formation constant for \(\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\) $$ K=\frac{\left[\mathrm{Mn}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{2}^{2-}\right]}{\left[\mathrm{Mn}^{2+}\right]\left[\mathrm{C}_{2} \mathrm{O}_{4}^{2-}\right]^{2}} $$

A solution is prepared by mixing \(50.0 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\) with \(50.0 \mathrm{~mL}\) of \(1.0 M \mathrm{KCl}\). Calculate the concentrations of \(\mathrm{Pb}^{2+}\) and \(\mathrm{Cl}^{-}\) at equilibrium. \(\left[K_{\mathrm{sp}}\right.\) for \(\mathrm{PbCl}_{2}(s)\) is \(\left.1.6 \times 10^{-5} .\right]\)
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