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Chapter 8: Problem 68

The solubility of \(\mathrm{CaCl}_{2}\) in water at \(25^{\circ} \mathrm{C}\) is \(81.1 \mathrm{g} / 100 \mathrm{mL}\) at \(0^{\circ} \mathrm{C}\) its solubility decreases to \(59.5 \mathrm{g} / 100 \mathrm{mL} .\) Answer the following questions about an aqueous solution of \(26.4 \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) in a volume of \(37.5 \mathrm{mL}\) a. At \(25^{\circ} \mathrm{C},\) is this a saturated solution? b. If the solution is cooled to \(0^{\circ} \mathrm{C},\) do you expect a precipitate to form? c. If the solution is slowly cooled to \(0^{\circ} \mathrm{C}\) and no precipitate forms, then what kind of solution is it?

Short Answer

Expert verified
Also, if no precipitate forms, what type of solution is it? Answer: The solution is not saturated at 25掳C. A precipitate will not form when the solution is cooled to 0掳C, and the type of solution is an unsaturated solution.

Step by step solution

01

(Step 1: Find the concentration of the given solution)

To find the concentration of the given solution, we need to divide the amount of \(\mathrm{CaCl}_2\) by the volume of the solution. In this case, we have \(26.4\, \mathrm{g}\) of \(\mathrm{CaCl}_2\) in \(37.5\, \mathrm{mL}\) of solution. So, the concentration is: \(\frac{26.4\, \mathrm{g}}{37.5\, \mathrm{mL}}\).
02

(Step 2: Determine if the solution is saturated at 25掳C)

At \(25^{\circ}\mathrm{C}\), the solubility of \(\mathrm{CaCl}_{2}\) is \(81.1\, \mathrm{g}/100\, \mathrm{mL}\). To determine if our solution is saturated at this temperature, we need to compare its concentration to the solubility. If the concentration is less than the solubility, then the solution is not saturated. We have: \(\frac{26.4\, \mathrm{g}}{37.5\, \mathrm{mL}} < \frac{81.1\, \mathrm{g}}{100\, \mathrm{mL}}\) a. This inequality holds, so at \(25^{\circ}\mathrm{C}\), the solution is not saturated.
03

(Step 3: Determine if a precipitate will form when cooled to 0掳C)

At \(0^{\circ}\mathrm{C}\), the solubility of \(\mathrm{CaCl}_{2}\) decreases to \(59.5\, \mathrm{g}/100\, \mathrm{mL}\). We need to compare the concentration of our solution to the solubility at this temperature. If the concentration is greater than the solubility, then a precipitate will form as the temperature is lowered. We have: \(\frac{26.4\, \mathrm{g}}{37.5\, \mathrm{mL}} > \frac{59.5\, \mathrm{g}}{100\, \mathrm{mL}}\) b. This inequality does not hold, so a precipitate will not form when the solution is cooled to \(0^{\circ}\mathrm{C}\).
04

(Step 4: Determine the type of solution if no precipitate forms at 0掳C)

If a solution is able to be cooled to \(0^{\circ}\mathrm{C}\) and no precipitate forms, this means that the concentration of the solute is still below its solubility at this temperature. Since the solution is not saturated, it is considered to be an unsaturated solution. c. Therefore, the type of solution is an unsaturated solution.

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Key Concepts

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

Saturated solution
A saturated solution is one in which the maximum amount of solute is dissolved in the solvent at a given temperature and pressure. This means that no more solute can be added without it starting to settle at the bottom as a solid. To determine if a solution is saturated, you compare the concentration of the solution to the solubility of the solute at that temperature.

For instance, at \(25^{\circ}\mathrm{C}\), the solubility of \(\mathrm{CaCl}_{2}\) is \(81.1\, \mathrm{g}/100\, \mathrm{mL}\). If you have a solution with a concentration less than this, like \(\frac{26.4\, \mathrm{g}}{37.5\, \mathrm{mL}}\), it means the solution is not saturated. Saturation is not reached, and thus, the solution can still dissolve additional \(\mathrm{CaCl}_{2}\) until it reaches that threshold.

Understanding whether a solution is saturated is essential in predicting how it will react when conditions such as temperature are changed.
Precipitate formation
Precipitate formation occurs when the concentration of a solute exceeds its solubility in a solution, leading to the formation of solid particles that settle out of the solution. This generally happens when a solution is cooled or when the solvent is evaporated. In the case of \(\mathrm{CaCl}_{2}\), at \(0^{\circ}\mathrm{C}\), the solubility significantly decreases to \(59.5\, \mathrm{g}/100\, \mathrm{mL}\). If our solution's concentration was higher than this value when cooled, a precipitate would form. However, in our scenario, the concentration \(\frac{26.4\, \mathrm{g}}{37.5\, \mathrm{mL}}\) remains below the solubility limit at \(0^{\circ}\mathrm{C}\), so no precipitate forms.

Precipitate formation is an indicator of a solution's saturation level and helps in determining the outcome of dilution or temperature changes in laboratory and industrial processes.
Unsaturated solution
An unsaturated solution is one that contains less solute than it can potentially dissolve at a certain temperature and pressure. This means additional solute can be added and will dissolve without changing the properties of the solution. In regards to \(\mathrm{CaCl}_{2}\), after cooling the solution to \(0^{\circ}\mathrm{C}\) and observing that no precipitate forms, it indicates that the solution is unsaturated.

In practical terms, if a solution is unsaturated, like the one described with \(26.4\, \mathrm{g}\) of \(\mathrm{CaCl}_{2}\) in \(37.5\, \mathrm{mL}\), it is capable of dissolving more \(\mathrm{CaCl}_{2}\) upon further addition.

This concept is foundational in understanding solution dynamics and is crucial in fields where solution preparation and concentration adjustments are required.

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Most popular questions from this chapter

When a solution of dithionite ions \(\left(\mathrm{S}_{2} \mathrm{O}_{4}^{2-}\right)\) is added to a solution of chromate ions \(\left(\mathrm{CrO}_{4}^{2-}\right),\) the products of the ensuing chemical reaction that occurs under basic conditions include soluble sulfite ions and solid chromium(III) hydroxide. This reaction is used to remove \(\mathrm{Cr}^{6+}\) from wastewater generated by factories that make chrome- plated metals. a. Write the net ionic equation for this redox reaction. b. Which element is oxidized and which is reduced? c. Identify the oxidizing and reducing agents in the reaction. d. How many grams of sodium dithionite would be needed to remove the \(\mathrm{Cr}^{6+}\) in \(100.0 \mathrm{L}\) of wastewater that contains \(0.00148 M\) chromate ion?

Zinc, copper, lead, and mercury ions are toxic to Atlantic salmon at concentrations of \(6.42 \times 10^{2} \mathrm{m} M, 7.16 \times\) \(10^{-3} \mathrm{m} M, 0.965 \mathrm{m} M,\) and \(5.00 \times 10^{-2} \mathrm{m} M,\) respectively. What are the corresponding concentrations in milligrams per liter?

Chlorine was first prepared in 1774 by heating a mixture of \(\mathrm{NaCl}\) and \(\mathrm{MnO}_{2}\) in sulfuric acid: $$\begin{aligned} \mathrm{NaCl}(a q)+\mathrm{H}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnO}_{2}(s) & \rightarrow \\ & \mathrm{Na}_{2} \mathrm{SO}_{4}(a q)+\mathrm{MnCl}_{2}(a q)+\mathrm{H}_{2} \mathrm{O}(\ell)+\mathrm{Cl}_{2}(g) \end{aligned}$$a. Assign oxidation numbers to the elements in each compound, and balance the redox reaction in acid solution. b. Write a net ionic equation describing the reaction for the formation of chlorine. c. If chlorine gas is inhaled, it causes pulmonary edema (fluid in the lungs) because it reacts with water in the alveolar sacs of the lungs to produce the strong acid \(\mathrm{HCl}\) and the weaker acid HOC1. Balance the equation for the conversion of \(\mathrm{Cl}_{2}\) to \(\mathrm{HCl}\) and \(\mathrm{HOCl}\).

Calcium nitrate is soluble, but calcium phosphate is not. Explain this difference in solubility

The concentration of manganese in one brand of soluble plant fertilizer is \(0.05 \%\) by mass. If a \(20 \mathrm{g}\) sample of the fertilizer is dissolved in 2.0 L of solution, what is the molarity of dissolved Mn in the solution?
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