Warning: foreach() argument must be of type array|object, bool given in /var/www/html/web/app/themes/studypress-core-theme/template-parts/header/mobile-offcanvas.php on line 20

Some lithium chloride, LiCl, is dissolved in 100 mL of water in one beaker and some \(\mathrm{Li}_{2} \mathrm{SO}_{4}\) is dissolved in 100 mL of water in another beaker. Both are at \(10^{\circ} \mathrm{C}\) and both are saturated solutions; some solid remains undissolved in each beaker. Describe what you would observe as the temperature is raised. The following data are available to you from a handbook of chemistry: $$\begin{array}{lll} \hline & \text { Solubility }(\mathrm{g} / 100 \mathrm{mL}) \\ \hline \text { Compound } & 10^{\circ} \mathrm{C} & 40^{\circ} \mathrm{C} \\ \hline \mathrm{Li}_{2} \mathrm{SO}_{4} & 35.5 & 33.7 \\\ \mathrm{LiCl} & 74.5 & 89.8 \\ \hline \end{array}$$

Short Answer

Expert verified
LiCl dissolves more at higher temp; Li₂SO₄ precipitates more.

Step by step solution

01

Initial Conditions

We begin with two separate beakers containing saturated solutions at 10°C. In one beaker, lithium chloride (LiCl) is dissolved, and in another, lithium sulfate (Li₂SO₄) is dissolved. Both solutions have some undissolved solids remaining at this temperature.
02

Understand Solubility Data

Consult the solubility data for each compound: - LiCl at 10°C: 74.5 g/100 mL - LiCl at 40°C: 89.8 g/100 mL - Li₂SO₄ at 10°C: 35.5 g/100 mL - Li₂SO₄ at 40°C: 33.7 g/100 mL This data indicates how much of each compound can maximally dissolve in 100 mL of water at specified temperatures.
03

Predict Changes for LiCl Solution

As the temperature increases to 40°C, the solubility of LiCl increases from 74.5 g/100 mL to 89.8 g/100 mL. Some of the previously undissolved LiCl will dissolve into the solution, decreasing the visible solid crystals and increasing the solution concentration.
04

Predict Changes for Li₂SO₄ Solution

For Li₂SO₄, the solubility decreases slightly from 35.5 g/100 mL at 10°C to 33.7 g/100 mL at 40°C. This means that some of the dissolved Li₂SO₄ may precipitate out of the solution, resulting in more solid forming, as temperature increases.
05

Observations on Temperature Rise

When the temperature is increased, the LiCl beaker will see a reduction in undissolved solids as they dissolve. Conversely, the Li₂SO₄ beaker will likely have more solids forming as precipitation occurs due to lower solubility at higher temperatures.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with Vaia!

Key Concepts

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

Lithium Chloride
Lithium chloride (LiCl) is a chemical compound known for its high solubility in water. At lower temperatures, such as 10°C, lithium chloride can dissolve up to 74.5 g in 100 mL of water. This is relatively high compared to many other salts. As the temperature of the solution increases, the solubility of LiCl also increases, allowing even more of the salt to dissolve. Specifically, at 40°C, LiCl can dissolve up to 89.8 g in the same amount of water.

When you heat a saturated solution of lithium chloride from 10°C to 40°C, more of the solid that was previously undissolved will enter the solution, reducing the solid residue. This demonstrates a direct temperature dependence where higher temperatures increase its ability to dissolve in water. If you observe a saturated solution heating up, you'll spot fewer solid particles as the temperature rises. This process highlights the strong temperature-solubility relationship lithium chloride has.
Lithium Sulfate
Lithium sulfate ( Li₂SO₄ ) behaves differently from lithium chloride when it comes to solubility. At 10°C, lithium sulfate can dissolve at a rate of 35.5 g per 100 mL of water. Unlike LiCl, as you increase the temperature to 40°C, the solubility of lithium sulfate slightly decreases to 33.7 g per 100 mL. This suggests that at higher temperatures, lithium sulfate tends not to dissolve as easily.

This can lead to more solid forming or precipitating out of the solution upon heating. As a result, if you were to observe a saturated solution of lithium sulfate while it's being warmed from 10°C to 40°C, you'd notice an increase in undissolved solid at the bottom of the container. This is because lithium sulfate exhibits an unusual inverse temperature-solubility relationship compared to many other salts.
Temperature Effects on Solubility
Solubility is significantly affected by temperature changes. For most salts, like lithium chloride, solubility increases with temperature. This means the solution can hold more dissolved solute at higher temperatures, causing previously undissolved solids to dissolve when the solution is heated. This is observed as a reduction in visible solid particles.

However, in some cases such as with lithium sulfate, solubility decreases with an increase in temperature. This is less common and is an interesting characteristic of certain compounds. When the temperature rises, the solubility decreases, meaning some of the solute will precipitate out and appear as more solid.
  • Direct relationship: solubility increases with temperature.
  • Inverse relationship: solubility decreases with temperature.
Thus, understanding temperature effects is key to predicting how a solution will behave when heated or cooled, indicating precipitation or further dissolution respectively.
Saturation and Precipitation
Saturation occurs when a solution can no longer dissolve additional solute at a given temperature and pressure, leaving undissolved solid. When a solution is saturated at a specific temperature, it contains the maximum concentration of solute possible.

If the temperature changes, this balance can be disrupted. For example, when a saturated lithium chloride solution is heated, its increased solubility allows more solute to dissolve, reducing saturation. Conversely, heating a saturated lithium sulfate solution will cause it to precipitate, forming more solid as excess solute comes out of the solution.

Precipitation is the result of solute concentration exceeding solubility, often due to temperature changes. As less solute can dissolve, the excess becomes a solid, making the solution appear cloudy or creating sediment. Understanding saturation and precipitation is crucial in predicting a compound's behavior within a saturated solution upon temperature variations.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Which of the following substances is likely to dissolve in water, and which is likely to dissolve in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right) ?\) (a) \(\mathrm{NaNO}_{3}\) (b) diethyl ether, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OCH}_{2} \mathrm{CH}_{3}\) (c) naphthalene, \(\mathrm{C}_{10} \mathrm{H}_{8}\) (d) \(\mathrm{NH}_{4} \mathrm{Cl}\)

Dimethylglyoxime [DMG, (CH \(\left._{3} \mathrm{CNOH} \text { ) }_{2}\right]\) is used as a reagent to precipitate nickel ion. Assume that \(53.0 \mathrm{g}\) of DMG has been dissolved in \(525 \mathrm{g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) (IMAGE CAN'T COPY) The red, insoluble compound formed between nickel(II) ion and dimethylglyoxime (DMG) is precipitated when DMG is added to a basic solution of \(\mathrm{Ni}^{2+}(\mathrm{aq})\) (a) What is the mole fraction of DMG? (b) What is the molality of the solution? (c) What is the vapor pressure of the ethanol over the solution at ethanol's normal boiling point of \(78.4^{\circ} \mathrm{C} ?\) (d) What is the boiling point of the solution? (DMG does not produce ions in solution.) \(\left(K_{\mathrm{bp}} \text { for ethanol }=\right.\) \(\left.+1.22^{\circ} \mathrm{C} / m\right)\)

The dispersed phase of a certain colloidal dispersion consists of spheres of diameter \(1.0 \times 10^{2} \mathrm{nm}\) (a) What is the volume \(\left(V=\frac{4}{3} \pi r^{3}\right)\) and surface area \((A=\) \(\left.4 \pi r^{2}\right)\) of each sphere? (b) How many spheres are required to give a total volume of \(1.0 \mathrm{cm}^{3} ?\) What is the total surface area of these spheres in square meters?

In chemical research we often send newly synthesized compounds to commercial laboratories for analysis. These laboratories determine the weight percent of \(\mathrm{C}\) and \(\mathrm{H}\) by burning the compound and collecting the evolved \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O} .\) They determine the molar mass by measuring the osmotic pressure of a solution of the compound. Calculate the empirical and molecular formulas of a compound, \(\mathrm{C}_{x} \mathrm{H}_{y}\) Cr, given the following information: (a) The compound contains \(73.94 \%\) C and \(8.27 \%\) H; the remainder is chromium. (b) At \(25^{\circ} \mathrm{C},\) the osmotic pressure of a solution containing 5.00 mg of the unknown dissolved in exactly 100 mL of chloroform solution is \(3.17 \mathrm{mm}\) Hg.

To make homemade ice cream, you cool the milk and cream by immersing the container in ice and a concentrated solution of rock salt (NaCl) in water. If you want to have a water-salt solution that freezes at \(-10 .^{\circ} \mathrm{C},\) what mass of NaCl must you add to \(3.0 \mathrm{kg}\) of water? (Assume the van't Hoff factor, \(i,\) for \(\mathrm{NaCl}\) is \(1.85 .\) )

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free