Working with solutions PDF

Preview

Summary

Download Working with solutions PDF

Description

Lesson 4.7 Working With Solutions Suggested Reading  Zumdahl Chapter 4 Section 4.3 Essential Question  What calculations are needed to make solutions in the lab?

Learning Objectives  Describe the molarity of solutions.  Perform calculations needed to make solutions.  Solve problems involving concentration, amount of solute and volume of solution. Reactions between two solid reactants often proceed very slowly or not at all. This is because the molecules or ions in a crystal tend to occupy approximately fixed positions, so that the chance of two molecules or ions coming together to react is small. For this reason, most reactions are run in liquid solutions. Reactant molecules are free to move throughout the liquid, and reaction is much faster. When you run reactions in liquid solutions, it is convenient to dispense the amounts of reactants by measuring out volumes of reactant solutions. In this lesson, we will discuss calculations involved in making solutions. Pay attention because moving forward you will be preparing your own solutions for use in your lab work when we are in need of solutions. Molar Concentration When we dissolve a substance in a liquid, we call the substance the solute and the liquid the solvent. For example, ammonia gas dissolves readily in water, and aqueous ammonia solutions are often used in the laboratory. In such solutions, ammonia gas is the solute and water is the solvent.

The general term concentration refers to the quantity of solute in a standard quantity of solution. Qualitatively, we say that a solution is dilute when the solute concentration is low and concentrated when the solute concentration is high. Usually these terms are used in a comparative sense and do not refer to a specific concentration. We say that one solution is more dilute or less concentrated than another. However, for commercially available solution, the term concentrated refers to the maximum concentration available. For example, concentrated aqueous ammonia contains about 28% NH3 by mass. In this example, the concentration of NH3 is expressed as the mass percent of solute (the mass of solute in 100g of solution). In the lab we usually express concentration in terms of molar concentration, or molarity (M), which is defined as the moles of solute dissolved in 1 liter (cubic decimeter) of solution. The mass-mole relationship makes molarity more convenient for working with laboratory reagents.

An aqueous solution that is 0.15 M NH3 (read this as "0.15 molar NH3") contains 0.15 mole NH3 per liter of solution. If you want to prepare a solution that is, for example, 0.200 M CuSO4, you place 0.200 mol of CuSO4 in a 1.000 L volumetric flask, or a proportional amount in a flask of a different size. You add a small quantity of water to dissolve the CuSO4. Then you fill the flask with with additional water to the mark on the neck and mix the solution. I typically pause and mix the solution when I have filled the flask half-way. As the solute dissolves it expands and occupies space in the solution. If you fill the volumetric flask to the mark before this expansion occurs you can end up with too much solvent and the concentration may be inaccurate as a result. Example: Calculating Molarity from Mass and Volume A sample of NaNO3, weighing 0.38 g is placed in a 50.0 mL volumetric flask. The flask is then filled with water to the mark of the neck. What is the molarity of the resulting solution? Solution: To calculate the molarity, you need the moles of solute. Therefore, you first covert grams NaNO3 to moles. The molarity equals the moles of solute divided by the liters of solution.

The last significant figure is underlined. The molarity is

One of the advantages of molarity as a concentration unit is that, for a solution of known molarity, the amount of solute is related to the volume of solution. Rather than having to weigh out a specified mass of substance, you can instead measure out a definite volume of solution of the substance, which is usually much easier. As the next example illustrates, molarity can be used as a factor for converting from moles of solute to liters of solution, and vice versa. Example: Using Molarity as a Conversion Factor An experiment calls for the addition to a reaction vessel of 0.184 g of sodium hydroxide, NaOH, in aqueous solution. How many millimeters of 0.150 M NaOH should be added? Solution: First you need to convert grams NaOH to moles NaOH, because molarity relates moles of solute to volume of solution. Then, you convert moles of NaOH to liters of solution, using molarity as a conversion factor. Here, 0.150 M means that 1 L of solution contains 0.150 moles of solute, so the conversion factor is:

Here is the calculation

You need to add 30.7 mL of a 0.150 M NaOH to the reaction vessel. You can see from the example that problems like these are solved like all other stoichiometry problems where dimensional analysis is critical to achieving the correct answer. Let the units guide you to the solution!

Diluting Solutions

Commercially available aqueous ammonia (28.0% NH3) is 14.8 M NH3. Suppose that you want as solution that is 1.00 M NH3. You need to dilute the concentrated solution with a definite quantity of water. For this purpose, you must know the relationship between the molarity of the solution before dilution (the initial molarity) and that after dilution (the final molarity). To obtain this relationship, first recall the equation defining molarity:

You can rearrange this to give Moles solute = molarity x liters of solution The product of molarity and the volume (in liters) gives the moles of solute in the solution. Writing Mi for the initial molar concentration and Vi for the initial volume of solution, you get Moles solute = Mi x Vi When the solution is diluted by adding water, the concentration and volume change to Mf and Vf giving

Moles solute = Mf x Vf Because the moles of solute has not changed during the dilution Mi x Vi = Mf x Vf This is a very useful equation! It is among the chemistry equations that I use most often. The next example illustrates how you can use this formula to find the volume of a concentrated solution needed to prepare a given volume of dilute solution.

Example: Diluting Solution You are given a solution of 14.8 M NH3. How many milliliters of this solution do you require to give 100.0 mL of 1.00 M NH3 when diluted. Solution: You know the final volume (100.0 mL), final concentration (1.00 M) and initial concentration (14.8 M) Write the dilution formula and rearrange it to give the initial volume.

Now substitute the known values into the equation

This result means that you need to add 6.76 mL to a 100 mL volumetric flask and then fill the mark with distilled water, pausing half way to mix....