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Chapter 1: Problem 16

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . How could the concentration of \(\mathrm{Sr}^{2+}\) ions in solution be decreased? (A) Adding some \(\operatorname{NaF}(s)\) to the beaker (B) Adding some \(\operatorname{Sr}\left(\mathrm{NO}_{3}\right)_{2}(s)\) to the beaker (C) By heating the solution in the beaker (D) By adding a small amount of water to the beaker, but not dissolving all the solid

Short Answer

Expert verified
The concentration of \(Sr^{2+}\) ions in solution can be decreased by adding some \(NaF(s)\) to the beaker.

Step by step solution

01

Identify the Equilibrium Shift

First, it is crucial to understand that by adding a reactive component to this system or changing its conditions, the system will adapt to maintain the solubility equilibrium. This is aligned with Le Chatelier's principle.
02

Examine Each Option

Now, evaluate each option individually to see how it would impact the system. (A) Adding some \(NaF(s)\) would increase the concentration of \(F^{-}\) ions in the solution. This in turn would push the equation to the left, reducing the concentration of \(Sr^{2+}\). (B) Adding \(Sr(NO_{3})_{2}(s)\) to the beaker would increase the concentration of \(Sr^{2+}\), shifting the equilibrium to the left but increasing the \(Sr^{2+}\) ion concentration. (C) Heating the solution would typically favor the endothermic direction. Without information regarding the enthalpy change (\(螖H\)), we cannot tell which direction would be favored by heat. (D) Adding water would increase the volume of solution, causing dilution. The concentration of ions would initially decrease, but the system would reestablish equilibrium by dissolving more \(SrF_{2}\), potentially leading to no real change in \(Sr^{2+}\) concentration.
03

Determine the Suitable Option

Based on the analysis, option (A) adding \(NaF(s)\) would effectively decrease the concentration of \(Sr^{2+}\) ions in solution without reestablishing equilibrium.

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Key Concepts

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

Molar Solubility
Molar solubility is a key concept when talking about the amount of a substance that can dissolve in a solution before the solution becomes saturated. Let's look at the example with the solubility of \( \text{SrF}_2 \). The molar solubility of \( \text{SrF}_2 \) at 298 K is given as \( 1.0 \times 10^{-3} \text{ M} \). This means that in one liter of water, \( 1.0 \times 10^{-3} \text{ moles} \) of \( \text{SrF}_2 \) can dissolve to reach saturation.

When additional solid is added to a saturated solution, it will not dissolve. That's because the system has reached the maximum amount that can dissolve at that temperature. Understanding molar solubility helps us predict how much of a substance can dissolve before the solution reaches equilibrium states.
Ion Concentration
Ion concentration refers to the amount of specific ions present in a solution. It is often measured in moles per liter (molarity). In the equilibrium of \( \text{SrF}_2 \), the ions we are interested in are \( \text{Sr}^{2+} \) and \( \ ext{F}^{-} \).

The concentration of \( \text{Sr}^{2+} \) ions can be decreased by affecting the equilibrium condition. According to the problem, adding \( \ ext{NaF}(s) \) would increase the \( \text{F}^{-} \) ion concentration, which according to Le Chatelier's Principle, would favor the reverse of the dissolution reaction, consequently decreasing the \( \text{Sr}^{2+} \) ion concentration.
Equilibrium Shift
The term equilibrium shift refers to changes in the equilibrium position of a chemical reaction in response to an external change. Le Chatelier鈥檚 Principle states that if a dynamic equilibrium is disturbed by changing the conditions, the position of equilibrium shifts to counteract the change.

In our case, if you introduce more \( \ ext{F}^{-} \) ions by adding \( \ ext{NaF}(s) \), the system shifts towards forming more solid \( \text{SrF}_2 \), thus decreasing the \( \text{Sr}^{2+} \) ions in solution. Shifts in equilibrium are crucial when we need to adjust the concentration of ions in chemical reactions, especially in industrial and laboratory settings.
Endothermic Reactions
Endothermic reactions absorb heat from the surroundings and are characterized by positive enthalpy changes (\( \Delta H \)). In our exercise, when considering heating the \( \text{SrF}_2 \) solution, it's important to remember heat typically favors the direction of an endothermic process.

However, without explicit information on whether the dissolution of \( \text{SrF}_2 \) is endothermic or exothermic, predicting the precise effect of heating remains speculative. In general, understanding whether a reaction is endothermic or exothermic helps anticipate how temperature changes might influence the position of the equilibrium.

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Most popular questions from this chapter

A student titrates 20.0 \(\mathrm{mL}\) of 1.0 \(M \mathrm{NaOH}\) with 2.0 \(\mathrm{M}\), \(\mathrm{HCO}_{2} \mathrm{H}\left(K_{\mathrm{a}}=1.8 \times 10^{-4}\right) .\) Formic acid is a monoprotic acid. If the formic acid were replaced with a strong acid such as HCl at the same concentration (2.0 M), how would that change the volume needed to reach the equivalence point? (A) The change would reduce the amount as the acid now fully dissociates. (B) The change would reduce the amount because the base will be more strongly attracted to the acid. (C) The change would increase the amount because the reaction will now go to completion instead of equilibrium. (D) Changing the strength of the acid will not change the volume needed to reach equivalence.

150 \(\mathrm{mL}\) of saturated \(\mathrm{SrF}_{2}\) solution is present in a 250 \(\mathrm{mL}\) beaker at room temperature. The molar solubility of \(\mathrm{SrF}_{2}\) at 298 \(\mathrm{K}\) is \(1.0 \times 10^{-3} \mathrm{M}\) . What are the concentrations of \(\mathrm{Sr}^{2+}\) and \(\mathrm{F}^{-}\) in the beaker? (A) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (B) \(\left[\mathrm{Sr}^{2+}\right]=1.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\) (C) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} M\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-3} M\) (D) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)

The following mechanism is proposed for a reaction: \(\begin{array}{ll}{2 \mathrm{A} \rightarrow \mathrm{B}} & {\text { (fast equilibrium) }} \\ {\mathrm{C}+\mathrm{B} \rightarrow \mathrm{D}} & {\text { (slow) }} \\ {\mathrm{D}+\mathrm{A} \rightarrow \mathrm{E}} & {\text { (fast) }}\end{array}\) Which of the following is the correct rate law for compete reaction? (A) Rate \(=k[\mathrm{C}]^{2}[\mathrm{B}]\) (B) Rate \(=k\left[\mathrm{Cl}[\mathrm{A}]^{2}\right.\) (C) Rate \(=k[\mathrm{C}][\mathrm{A}]^{3}\) (D) Rate \(=k[\mathrm{D}][\mathrm{A}]\)

Use the following information to answer questions 29-31. Pennies are made primarily of zinc, which is coated with a thin layer of copper through electroplating, using a setup like the one above. The solution in the beaker is a strong acid (which produces H' ions), and the cell is wired so that the copper electrode is the anode and zinc penny is the cathode. Use the following reduction potentials to answer questions \(29-31 .\) $$\begin{array}{|l|l|}\hline \text { Half-Reaction } & {\text { Standard Reduction Potential }} \\ \hline \mathrm{Cu}^{2++2 e^{-} \rightarrow \mathrm{Cu}(s)} & {+0.34 \mathrm{V}} \\ \hline 2 \mathrm{H}^{++2 e^{-} \rightarrow \mathrm{H}_{2}(g)} & {0.00 \mathrm{V}} \\ \hline \mathrm{Ni}^{2++2 e^{-} \rightarrow \mathrm{Ni}(s)} & {-0.25 \mathrm{V}} \\\ \hline \mathrm{Zn}^{2++2 e^{-} \rightarrow \mathrm{Zn}(s)} & {-0.76 \mathrm{V}} \\ \hline\end{array}$$ When the cell is connected, which of the following reactions takes place at the anode? (A) \(\mathrm{Cu}^{2+}+2 e^{-} \rightarrow \mathrm{Cu}(\mathrm{s})\) (B) \(\mathrm{Cu}(s) \rightarrow \mathrm{Cu}^{2+}+2 e^{-}\) (C) \(2 \mathrm{H}^{+}+2 e^{-} \rightarrow \mathrm{H}_{2}(g)\) (D) \(\mathrm{H}_{2}(g) \rightarrow 2 \mathrm{H}^{+}+2 e^{-}\)

Consider the following reaction showing photosynthesis: $$\begin{array}{c}{6 \mathrm{CO}_{2}(g)+6 \mathrm{H}_{2} \mathrm{O}(l) \rightarrow \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}(s)+6 \mathrm{O}_{2}(g)} \\ {\Delta H=+2800 \mathrm{kJ} / \mathrm{mol}}\end{array}$$ Which of the following is true regarding the thermal energy in this system? (A) It is transferred from the surroundings to the reaction. (B) It is transferred from the reaction to the surroundings. (C) It is transferred from the reactants to the products. (D) It is transferred from the products to the reactants.
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