Solubility Curve Practice Problems Worksheet 1 Answer Key

The solubility curve is a graphical representation that illustrates the relationship between temperature and the maximum amount of solute that can dissolve in a given amount of solvent. Understanding and interpreting solubility curves is a fundamental skill in chemistry, with wide-ranging applications from pharmaceutical development to food science. To master this skill, solving practice problems is essential. This article provides an in-depth exploration of solubility curves, complete with worked examples and an answer key to enhance your understanding.

Understanding Solubility Curves: A Comprehensive Guide

Solubility curves are essential tools in chemistry for visualizing the relationship between temperature and the solubility of a substance. Before diving into practice problems, let's cover the basics.

What is a Solubility Curve?

A solubility curve is a graph that plots the solubility of a substance (usually a salt) as a function of temperature. Solubility is typically expressed as grams of solute per 100 grams of solvent (usually water). The curve shows the maximum amount of solute that can dissolve in the solvent at different temperatures, creating a saturated solution.

Key Components of a Solubility Curve

• X-axis: Represents the temperature, usually in degrees Celsius (°C).
• Y-axis: Represents the solubility, usually in grams of solute per 100 grams of water (g/100g H₂O).
• The Curve: The line on the graph represents the solubility of the solute at different temperatures. Any point on the curve indicates a saturated solution.
• Points Above the Curve: Represent supersaturated solutions, which contain more solute than can normally dissolve at that temperature.
• Points Below the Curve: Represent unsaturated solutions, which contain less solute than the maximum amount that can dissolve at that temperature.

Interpreting a Solubility Curve

• Increase in Solubility with Temperature: For most solid solutes, solubility increases with temperature. This means that as the temperature rises, more solute can dissolve in the solvent.
• Decrease in Solubility with Temperature: Some solutes, particularly gases, exhibit a decrease in solubility as temperature increases.
• Identifying Saturated, Unsaturated, and Supersaturated Solutions:
  • Saturated Solution: Lies exactly on the curve.
  • Unsaturated Solution: Lies below the curve.
  • Supersaturated Solution: Lies above the curve.
• Determining Solubility at a Specific Temperature: Locate the desired temperature on the x-axis, trace vertically to the curve, and then trace horizontally to the y-axis to read the solubility value.
• Determining the Temperature for a Specific Solubility: Locate the desired solubility on the y-axis, trace horizontally to the curve, and then trace vertically to the x-axis to read the temperature value.

Solubility Curve Practice Problems

Now, let's dive into a series of practice problems to reinforce your understanding of solubility curves. Each problem will be presented, followed by a step-by-step solution.

Problem 1:

Using the solubility curve provided (see a hypothetical solubility curve below), determine the solubility of potassium nitrate (KNO₃) at 50°C.

Hypothetical Solubility Curve (Example):

Imagine a graph where:

• At 0°C, KNO₃ solubility is 15 g/100g H₂O.
• At 20°C, KNO₃ solubility is 30 g/100g H₂O.
• At 40°C, KNO₃ solubility is 60 g/100g H₂O.
• At 60°C, KNO₃ solubility is 105 g/100g H₂O.
• At 80°C, KNO₃ solubility is 160 g/100g H₂O.

Solution:

1. Locate the Temperature: Find 50°C on the x-axis.
2. Trace Vertically: Trace a vertical line upwards from 50°C until it intersects the KNO₃ solubility curve.
3. Trace Horizontally: From the point of intersection, trace a horizontal line to the y-axis.
4. Read the Solubility: Read the value on the y-axis. In this hypothetical curve, you would need to interpolate. Since 50°C is halfway between 40°C and 60°C, the solubility is approximately halfway between 60 g/100g H₂O and 105 g/100g H₂O. This gives us approximately 82.5 g/100g H₂O.

Answer: The solubility of KNO₃ at 50°C is approximately 82.5 g/100g H₂O.

Problem 2:

A solution contains 70 g of sodium chloride (NaCl) in 100 g of water. At what temperature would this solution be saturated according to the solubility curve (assume the curve is as described below)?

Hypothetical Solubility Curve (Example):

• At 0°C, NaCl solubility is 35 g/100g H₂O.
• At 20°C, NaCl solubility is 36 g/100g H₂O.
• At 40°C, NaCl solubility is 37 g/100g H₂O.
• At 60°C, NaCl solubility is 38 g/100g H₂O.
• At 80°C, NaCl solubility is 39 g/100g H₂O.
• At 100°C, NaCl solubility is 40 g/100g H₂O.

Note: NaCl's solubility increases very slowly with temperature.

Solution:

1. Locate the Solubility: Find 70 g/100g H₂O on the y-axis. In this example, 70g is far above any solubility point on the curve. This implies one of two things: either the provided curve is inaccurate, or a much higher temperature is required to dissolve 70g NaCl in 100g water.
2. Trace Horizontally: Trace a horizontal line from 70 g/100g H₂O until it intersects the NaCl solubility curve.
3. Trace Vertically: From the point of intersection, trace a vertical line to the x-axis.
4. Read the Temperature: Read the value on the x-axis. Since 70g is far beyond the maximum solubility shown on the graph, we can infer that a very high temperature (likely significantly above 100°C) would be needed.

Answer: Based on the limited solubility curve data provided, we cannot accurately determine the temperature. The temperature would be significantly higher than 100°C, and outside the range of the hypothetical graph. It highlights how NaCl solubility is relatively unaffected by temperature compared to other salts like KNO₃.

Problem 3:

At 30°C, the solubility of ammonium chloride (NH₄Cl) is 40 g/100g H₂O. If you dissolve 30 g of NH₄Cl in 100 g of water at 30°C, is the solution saturated, unsaturated, or supersaturated?

Solution:

1. Compare the Actual Amount Dissolved with the Solubility: The solubility of NH₄Cl at 30°C is 40 g/100g H₂O. You have dissolved 30 g/100g H₂O.
2. Determine the Solution Type:
   • Since 30 g/100g H₂O is less than 40 g/100g H₂O, the solution is unsaturated.

Answer: The solution is unsaturated.

Problem 4:

You have a saturated solution of potassium chlorate (KClO₃) in 200 g of water at 70°C. According to the solubility curve (provided below), how much KClO₃ is dissolved in the solution?

Hypothetical Solubility Curve (Example):

• At 50°C, KClO₃ solubility is 20 g/100g H₂O.
• At 60°C, KClO₃ solubility is 25 g/100g H₂O.
• At 70°C, KClO₃ solubility is 38 g/100g H₂O.
• At 80°C, KClO₃ solubility is 47 g/100g H₂O.
• At 90°C, KClO₃ solubility is 58 g/100g H₂O.

Solution:

1. Find the Solubility at the Given Temperature: At 70°C, the solubility of KClO₃ is 38 g/100g H₂O.

2. Adjust for the Amount of Water: The solubility is given for 100 g of water, but you have 200 g of water. To find the amount dissolved in 200 g of water, multiply the solubility by 2.

   • 38 g KClO₃ / 100 g H₂O * 2 = 76 g KClO₃ / 200 g H₂O

Answer: 76 g of KClO₃ is dissolved in the solution.

Problem 5:

A solution of lead(II) nitrate (Pb(NO₃)₂) contains 80 g of the salt in 100 g of water at 40°C. Is this solution saturated, unsaturated, or supersaturated, given the solubility curve information below?

Hypothetical Solubility Curve (Example):

• At 20°C, Pb(NO₃)₂ solubility is 55 g/100g H₂O.
• At 30°C, Pb(NO₃)₂ solubility is 68 g/100g H₂O.
• At 40°C, Pb(NO₃)₂ solubility is 80 g/100g H₂O.
• At 50°C, Pb(NO₃)₂ solubility is 95 g/100g H₂O.
• At 60°C, Pb(NO₃)₂ solubility is 115 g/100g H₂O.

Solution:

1. Find the Solubility at the Given Temperature: At 40°C, the solubility of Pb(NO₃)₂ is 80 g/100g H₂O.

2. Compare the Actual Amount Dissolved with the Solubility: You have 80 g of Pb(NO₃)₂ dissolved in 100 g of water.

   • Since the amount dissolved (80 g) is equal to the solubility at that temperature (80 g/100g H₂O), the solution is saturated.

Answer: The solution is saturated.

Problem 6:

You cool a saturated solution of copper(II) sulfate (CuSO₄) from 90°C to 30°C. If you started with 100 g of water, how much CuSO₄ will precipitate out of the solution, based on the curve below?

Hypothetical Solubility Curve (Example):

• At 30°C, CuSO₄ solubility is 25 g/100g H₂O.
• At 40°C, CuSO₄ solubility is 30 g/100g H₂O.
• At 50°C, CuSO₄ solubility is 35 g/100g H₂O.
• At 60°C, CuSO₄ solubility is 45 g/100g H₂O.
• At 70°C, CuSO₄ solubility is 55 g/100g H₂O.
• At 80°C, CuSO₄ solubility is 70 g/100g H₂O.
• At 90°C, CuSO₄ solubility is 80 g/100g H₂O.

Solution:

1. Find the Solubility at the Initial Temperature: At 90°C, the solubility of CuSO₄ is 80 g/100g H₂O.

2. Find the Solubility at the Final Temperature: At 30°C, the solubility of CuSO₄ is 25 g/100g H₂O.

3. Calculate the Difference in Solubility: Subtract the solubility at the final temperature from the solubility at the initial temperature.

   • 80 g/100g H₂O - 25 g/100g H₂O = 55 g/100g H₂O

Answer: 55 g of CuSO₄ will precipitate out of the solution.

Problem 7:

At what temperature do potassium nitrate (KNO₃) and sodium nitrate (NaNO₃) have the same solubility, given the data below?

Hypothetical Solubility Curve (Example):

• At 20°C: KNO₃ solubility is 30 g/100g H₂O; NaNO₃ solubility is 88 g/100g H₂O.
• At 40°C: KNO₃ solubility is 60 g/100g H₂O; NaNO₃ solubility is 104 g/100g H₂O.
• At 60°C: KNO₃ solubility is 105 g/100g H₂O; NaNO₃ solubility is 122 g/100g H₂O.
• At 80°C: KNO₃ solubility is 160 g/100g H₂O; NaNO₃ solubility is 148 g/100g H₂O.
• At 100°C: KNO₃ solubility is 245 g/100g H₂O; NaNO₃ solubility is 170 g/100g H₂O.

Solution:

1. Analyze the Data: Look for the point where the solubility values for KNO₃ and NaNO₃ are closest or equal.

2. Identify the Temperature: Based on the hypothetical data, the solubilities intersect somewhere between 70°C and 90°C. Because the solubility of KNO₃ increases more rapidly than NaNO₃, and because NaNO₃'s solubility begins to plateau and even slightly decrease relative to KNO₃, we can estimate the intersection point.

   Interpolating between 60°C and 80°C is necessary.

   • At 60°C, the difference is 122-105 = 17 g/100g H₂O
   • At 80°C, the difference is 160-148 = 12 g/100g H₂O

   We can approximate that the solubilities would intersect somewhere between these two data points. Since the difference decreases by 5 g between 60 and 80 degrees, we can very roughly approximate that the intersection point is 2/5ths of the way between 60 and 80 degrees. 2/5ths of 20 degrees is 8 degrees. Therefore, the approximate temperature is 68°C. A more precise answer would require either a higher resolution graph or the exact equation defining both curves.

Answer: Approximately 68°C.

Problem 8:

You have 50 g of water and want to make a saturated solution of silver nitrate (AgNO₃) at 60°C. How much AgNO₃ do you need, based on the solubility curve information below?

Hypothetical Solubility Curve (Example):

• At 40°C, AgNO₃ solubility is 310 g/100g H₂O.
• At 50°C, AgNO₃ solubility is 380 g/100g H₂O.
• At 60°C, AgNO₃ solubility is 455 g/100g H₂O.
• At 70°C, AgNO₃ solubility is 525 g/100g H₂O.
• At 80°C, AgNO₃ solubility is 590 g/100g H₂O.

Solution:

1. Find the Solubility at the Given Temperature: At 60°C, the solubility of AgNO₃ is 455 g/100g H₂O.

2. Adjust for the Amount of Water: The solubility is given for 100 g of water, but you have 50 g of water. To find the amount needed for 50 g of water, divide the solubility by 2.

   • 455 g AgNO₃ / 100 g H₂O / 2 = 227.5 g AgNO₃ / 50 g H₂O

Answer: You need 227.5 g of AgNO₃.

Problem 9:

A solution contains 60 g of solute X in 100 g of water at 50°C. According to the solubility curve (provided below), if the solution is cooled to 20°C, how much solute X will precipitate out?

Hypothetical Solubility Curve (Example):

• At 20°C, solute X solubility is 20 g/100g H₂O.
• At 30°C, solute X solubility is 35 g/100g H₂O.
• At 40°C, solute X solubility is 48 g/100g H₂O.
• At 50°C, solute X solubility is 60 g/100g H₂O.
• At 60°C, solute X solubility is 73 g/100g H₂O.

Solution:

1. Find the Solubility at the Initial Temperature: At 50°C, the solubility of solute X is 60 g/100g H₂O. This means the initial solution is saturated.

2. Find the Solubility at the Final Temperature: At 20°C, the solubility of solute X is 20 g/100g H₂O.

3. Calculate the Difference in Solubility: Subtract the solubility at the final temperature from the solubility at the initial temperature.

   • 60 g/100g H₂O - 20 g/100g H₂O = 40 g/100g H₂O

Answer: 40 g of solute X will precipitate out.

Problem 10:

You have a saturated solution of a certain salt in 250 g of water at 80°C. When the solution is cooled to 30°C, 65 grams of the salt crystallize out. Use this information to determine the solubility of the salt at 30°C, assuming that at 80°C the solubility of the salt is 91g/100g H₂O.

Solution:

1. Calculate total salt dissolved at 80°C:

   Since solubility at 80°C is 91 g/100g H₂O and you have 250g of H₂O, calculate total grams of salt dissolved in 250g H₂O at 80°C:

   91g salt / 100g H₂O * 250g H₂O = 227.5g salt

2. Calculate the amount of salt left at 30°C:

   If 65g crystallized after cooling to 30°C, that means:

   227.5g - 65g = 162.5g of salt are still dissolved in solution at 30°C.

3. Solubility calculation:

   Now that you know 162.5g salt are dissolved in 250g H₂O at 30°C, you can calculate grams per 100g H₂O, which is the definition of solubility:

   162.5g / 250g * 100g = 65g/100g H₂O

Answer: The solubility of the salt at 30°C is 65g/100g H₂O

Factors Affecting Solubility

While solubility curves primarily depict the relationship between temperature and solubility, other factors can influence how much solute can dissolve in a solvent. These include:

• Pressure: Primarily affects the solubility of gases. Henry's Law states that the solubility of a gas in a liquid is directly proportional to the pressure of the gas above the liquid.
• Nature of Solute and Solvent: The "like dissolves like" principle suggests that polar solvents dissolve polar solutes, and nonpolar solvents dissolve nonpolar solutes.
• Presence of Other Solutes: The presence of other dissolved substances can sometimes increase or decrease the solubility of a particular solute, although this is usually a minor effect.

Practical Applications of Solubility Curves

Solubility curves are more than just theoretical constructs. They have numerous practical applications across various fields:

• Crystallization: Used in chemical industries to purify substances by controlling the temperature and solubility to induce crystallization.
• Pharmaceuticals: Essential for formulating drugs, ensuring the right concentration and solubility for effective delivery.
• Food Science: Helps in understanding the solubility of sugars, salts, and other ingredients in food processing.
• Environmental Science: Used to study the solubility of pollutants in water and soil.

Conclusion

Mastering the interpretation and application of solubility curves is a crucial skill in chemistry. By working through practice problems, you can develop a solid understanding of the relationships between temperature, solubility, and solution types. Understanding how to analyze solubility curves will provide you with a strong foundation for more advanced concepts in chemistry and related fields. Keep practicing and exploring different scenarios to enhance your expertise.