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Chapter 1: Problem 14

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) \(\left[\mathrm{Sr}^{2+}\right]=2.0 \times 10^{-3} \mathrm{M}\left[\mathrm{F}^{-}\right]=2.0 \times 10^{-3} M\)

Short Answer

Expert verified
The correct answer is (B) \( [Sr^{2+}] = 1.0 \times 10^{-3} M, [F^-] = 2.0 \times 10^{-3} M \).

Step by step solution

01

Write the Dissolution Reaction

Start by writing down the dissolution reaction of \( SrF_2 \). Here, it is clear that 1 mole of \( SrF_2 \) dissociates into 1 mole of \( Sr^{2+} \) ions and 2 moles of \( F^- \) ions. The equation hence becomes: \( SrF_2 \longrightarrow Sr^{2+} + 2F^- \).
02

Use Stoichiometry to Determine Ion Concentrations

Using the stoichiometry of the equation, we know that for every 1 mole of \( SrF_2 \) that dissolves, we have 1 mole of \( Sr^{2+} \) ion and 2 moles of \( F^- \) ion. Given the molar solubility of \( SrF_2 \) as \( 1.0 \times 10^{-3} M \), the concentration of \( Sr^{2+} \) will thus be \( 1.0 \times 10^{-3} M \) and the concentration of \( F^- \) will be \( 2.0 \times 10^{-3} M \). The ratio of the concentrations draws directly from the coefficients in the balanced chemical equation.
03

Match the Ion Concentrations with the Answer Choices

The determination made in step 2 matches with answer choice B. Thus, B is the correct answer.

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Most popular questions from this chapter

Use the following information to answer questions 25-28. A voltaic cell is created using the following half-cells: \(\begin{array}{ll}{\mathrm{Cr}^{3+}+3 e \rightarrow \mathrm{Cr}(s)} & {E^{\circ}=-0.41 \mathrm{V}} \\ {\mathrm{Pb}^{2+}+2 e \rightarrow \mathrm{Pb}(s)} & {E^{\circ}=-0.12 \mathrm{V}}\end{array}\) The concentrations of the solutions in each half-cell are 1.0 M. Which of the following occurs at the cathode? (A) \(\mathrm{Cr}^{3+}\) is reduced to \(\mathrm{Cr}(\mathrm{s})\) (B) \(\mathrm{Pb}^{2+}\) is reduced to \(\mathrm{Pb}(\mathrm{s})\) (C) \(\mathrm{Cr}(s)\) is oxidized to \(\mathrm{Cr}^{3+}\) (D) \(\quad \mathrm{Pb}(s)\) is oxidized to \(\mathrm{Pb}^{2+}\)

A 2.0 L flask holds 0.40 g of helium gas. If the helium is evacuated into a larger container while the temperature is held constant, what will the effect on the entropy of the helium be? (A) It will remain constant because the number of helium molecules does not change. (B) It will decrease because the gas will be more ordered in the larger flask. (C) It will decrease because the molecules will collide with the sides of the larger flask less often than they did in the smaller flask. (D) It will increase because the gas molecules will be more dispersed in the larger flask.

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 \(1.0 \times 10^{-3} \mathrm{M}\) . If some of the solution evaporates overnight, which of the following will occur? (A) The mass of the solid and the concentration of the ions will stay the same. (B) The mass of the solid and the concentration of the ions will increase. (C) The mass of the solid will decrease, and the concentration of the ions will stay the same. (D) The mass of the solid will increase, and the concentration of the ions will stay the same.

When calcium chloride \(\left(\mathrm{CaCl}_{2}\right)\) dissolves in water, the temperature of the water increases dramatically. Compared to \(\mathrm{CaCl}_{2},\) what must be true regarding the hydration energy of \(\mathrm{CaF}_{2} ?\) (A) It would be greater because fluoride is smaller than chloride. (B) It would be the same because the charges of fluoride and chloride are identical. (C) It would be the same because hydration energy is only dependent on the IMFs present in water. (D) It would be smaller because the molar mass of \(\mathrm{CaF}_{2}\) is smaller than that of \(\mathrm{CaCl}_{2}\) .

1.50 g of \(\mathrm{NaNO}_{3}\) is dissolved into 25.0 \(\mathrm{mL}\) of water, causing the temperature to increase by \(2.2^{\circ} \mathrm{C}\) . The density of the final solution is found to be 1.02 \(\mathrm{g} / \mathrm{mL}\) . Which of the following expressions will correctly calculate the heat gained by the water as the NaNO, dissolves? Assume the volume of the solution remains unchanged. (A) \((25.0)(4.18)(2.2)\) (B) \(\frac{(26.5)(4.18)(2.2)}{1.02}\) (C) \(\frac{(1.02)(4.18)(2.2)}{1.50}\) (D) \((25.0)(1.02)(4.18)(2.2)\)
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