Unit 26: Solution Chemistry

 

Unit Overview:

In the last few units, you defined solutions, explored solubility, described solutions, and calculated solution concentration.

 

In this unit, you will apply your knowledge of solution concentration to explore solution stoichiometry problem-solving.

 

How do you make a solution of known concentration?

In order to make a solution of a specific concentration, follow these steps:

1.    Calculate the mass of solute required. You will need to complete a molarity problem to calculate this.  First, use the given molarity as a conversion factor to convert the volume of solution to moles of solute.  Then, use the molar mass of the as a conversion factor to convert the moles to the mass of solute. [So, you may want to have your  Interactive Periodic Table pulled up so that you can access it for molar mass values.]

2.    Carefully, mass out the calculated amount of solute. In the lab, you use a balance to carefully measure the required amount of solute. 

3.    Carefully measure the volume of water required to make the solution. In the lab, use a graduated cylinder (or a volumetric flask) to measure the exact volume of water to make the solution.

4.    Dissolve the solute in the solvent. Dissolve the solute in the solvent to make the exact volume of the exact molarity.

 

Example 1:  Describe how you would make 250.0 mL of a 1.25 M NaCl solution.

Step 1:  Calculate the mass of solute required.

Start with the volume of a solution:

First, use molarity as a conversion factor to convert to moles:

Then, use molar mass as a conversion factor to convert to mass:

To find the mass of solute required!

 

250.0 mL

1.25 mol NaCl

1000 mL

58.44 g NaCl

 mol NaCl

 

= 18.3 g NaCl

Step 2:  Carefully mass out the calculated amount of solute.

Use a balance to carefully measure 18.3 g of NaCl.

 

Use a graduated cylinder to measure 250 mL volume of water into a beaker.

 

 

 

 

Step 3:  Carefully measure the volume of water required to make the solution.

 

Or, use a volumetric flask to measure exactly 250 mL of water.

 

 

 

Step 4:  Dissolve the solute in the solvent.

Use a stirring rod to dissolve the salt in the water.

 

 

Watch the following video to help further you visualize and understand how to make a solution of known concentration:

Solution Preparation

 

 

Practice 1:  Determine the mass of solute that would be required to make each of the following solutions.

1.   40 mL of 0.7 M KBr

2.   300 mL of 2.5 M HCl

3.   70 mL of 0.5 M BaS

                                        Answer Key

 

How do you dilute a solution?

Solutions are generally described as concentrated or dilute. A concentrated solution describes a solution that has a relatively larger amount of solute dissolved in the solvent.  A dilute solution describes a solution that has a relatively smaller amount of solute dissolved in the solvent.

 

In a microscopic representation of the two general categories of solutions, a concentrated solution has a greater number of solute particles between the water molecules.  A dilute solution has a smaller number of solute particles between the water molecules.

 

 

A concentrated solution can be diluted by adding water to it.  To determine how to dilute a solution, the dilution equation is utilized.  

 

The dilution equation: VcMc = VdMd

Where:

Vc = volume of the concentrated solution 

Mc = molarity of the concentrated solution

Vd = volume of the dilute solution

Md = molarity of the dilute solution

 

To solve dilution problems, carefully solve the “missing variable.”

 

Example 2: If the solution prepared in the above example 1 is diluted to 750 mL, what is the concentration of the dilute solution?

 

To solve this problem, use the dilution equation:

VcMc = VdMd

 

The question asks me to calculate the molarity of the dilute solution, so that is the “missing variable,”

VcMc = VdMd

 

The other 3 variables are known.  The solution from example 1 is the concentrated solution, so those values are placed into the equation.

(250 mL)(1.25 M) = VdMd

 

We are told that the solution is diluted to 750 mL, so that is the volume of the dilute solution.

(250 mL)(1.25 M) = (750 mL)Md

 

Then, solve for the missing variable:

0.42 M = Md

 

Watch the following video to help further you understand how to dilute a solution to a specific concentration:

Solution Dilution

 

 

Practice 2:  Complete the practice problems in this online quiz.

 

How are reactions of solutions mathematically analyzed?

In unit 22, you explored stoichiometry as a method to analyze the relative amounts of reactants and products involved in a chemical reaction.  In this unit, you will expand your stoichiometry skills to include the use of concentration within this analysis. 

 

When reactions include solutions, molarity is used in the stoichiometry problem-solving process to relate the volume of solution to the moles of reactant. Molarity is an additional conversion factor that is used with molar mass and coefficients.

 

Example 3: If the diluted solution from the above example 2 reacts with an excess of silver nitrate, how many grams of silver chloride will precipitate?

NaCl (aq)   +   AgNO3 (aq)   →   AgCl (s)   +   NaNO3 (aq)

 

 

Start with the given substance:

Convert the volume to moles, using the molarity as a conversion factor:

Relate moles of reactant to moles of product with coefficients:

Convert the moles to mass, using molar mass:

To calculate the mass of precipitate!

750 mL

0.42 mol NaCl

1000 mL

 

 

1 mol AgCl

1 mol NaCl

 

143.32 g AgCl

1 mol AgCl

 

 

= 45.1 g AgCl

Watch the following video to help you connect your understandings of solutions to your understandings of stoichiometry:

Solution Stoichiometry

 

 

Practice 3: Complete the practice problems.

 

Solve each of the following solution stoichiometry problems:

1.) If 25.0 mL of 0.350 M NaOH is added to a copper (II) sulfate solution, how many grams of copper (II) hydroxide will precipitate?

2 NaOH (aq)   +   CuSO4 (aq)    →   Cu(OH)2 (s)   +   Na2SO4 (aq)

 

2.) How many mL of 0.715 M HCl is required to completely react with 1.25 grams of sodium carbonate?

2 HCl (aq)   +   Na2CO3 (s)   →   2 NaCl (aq)   +   H2O (l)    +  CO2 (g)

 

3.) How many mL of 0.280 M barium nitrate are required to completely react with 25.0 mL of 0.350 M aluminum sulfate?

3 Ba(NO3)2 (aq)   +   Al2(SO4)3 (aq)  →   3 BaSO4(s)  +  2 Al(NO3)3 (aq)

 

                                                                                                                                Answer Key

 

                                                                                                                    

 

ChemLab: Concentration Phet Simulation

Overview:

When describing a solution, its concentration indicates the relative amounts of solute and solvent.  Concentration increases as the amount of solute in a fixed amount of solvent increases.  In this lab, you will explore concentration by examining the relative amounts of solute and solvent.

 

Embed the Phet Simulation: Concentration

Directions:

1.       Download the Student Exploration Sheet.

2.       Practice using the Phet simulation.

3.      Follow the instructions in the Exploration Sheet to explore the relationships between the amount of solute and solvent.  Be sure to record your answers, so that you can upload your completed lab sheet.