Unit 15 s and Molarity INTRODUCTION In addition to chemical equations chemists and chemistry students encounter homogeneous mixtures or solutions quite frequently. s are the practical means to deliver certain reagents or media needed for a particular chemical reaction. A chemist cannot help but use solutions. Therefore, a thorough understanding of their general properties, descriptions and preparations is essential for further work in chemistry. OBJECTIVES 1. The student will describe or identify the general types and properties of solutions as described in this unit. 2. The student will calculate the molarity of various solutions using the definition of molarity as described in this unit. 3. The student will describe and perform the calculations for the preparation of solutions including methods that include use of concentrated solutions. DISCUSSION A. A solution is a homogeneous mixture of two or more substances. The substances are most often compounds. The term homogeneous describes a system of only one phase. The possible phases are solid, liquid and gas. The solution we will encounter will be those in the liquid phase. A solution contains at least one solute and a solvent. Generally there is only one solute. The solute is the compound that is dissolved in the solvent. The solvent is always present in a greater amount than the solute. The most common solutions which include those that will work with have water as a solvant. Common solutes are the compounds HCI (Hydrochloric acid), HNO 3 (Nitric acid), H 2 SO 4 (Sulfuric acid), HC 2 H 3 O 2 (Acetic acid), NH 4 OH (Ammonium Hydroxide or Ammonia water) and NaOH (Sodium Hydroxide). These compounds mix with water to form solutions and are the common solutions found in almost all chemistry laboratories. Some of these compounds are solids, some are liquids and one, HCl, is a gas but they dissolved in water to produce aqueous solutions.
B. Properties of s The above mentioned solutions and any true solution will exhibit the following general properties. 1. s are homogeneous. They appear as only one phase, but may contain others dissolved in it. 2. s share variable concentrations. They can have any ratio of solute to solvent. They are not compounds! 3. s will not settle out upon standing. The solute in a solvent will not settle to the bottom of the container no matter how long one waits. 4. The compounds of a solution can be separated. The solute can be recovered from the solution in many cases by evaporating the solvent, water. C. Molarity The molarity of a solution describes quantitatively the amount of solute present in a given volume of solution. The molarity (M) is the number of moles of solute per liter of solution. Molarity (M) = no. moles of solute liter of solution This is the most often used unit of concentration in chemistry and as such will be the focus of our intentions in this unit. D. Preparation of s and Calculations s are prepared in two ways. One method is to mix a measured amount of solute with a small amount of solvent and then add solvent until the desired volume of the solution is reached. The solute does take up some volume in the solution. The other method is to add water to a volume of an already prepared solution until the solution has been diluted to the desired concentration. This is the most common method because the compounds mentioned are often delivered as very concentrated solutions. s made from these concentrated solutions can be dilute or concentrated. A solution with a molarity of 6 or a 6 molar solutions is called dilute. While 12-16 molar solutions are called concentrated.
Example Problem (1) How would you prepare 500 mls of a 3 molar solution of NaOH? Step 1 - Given 500 mls of 3M NaOH Step 2 To find How to prepare it or how many moles or grams of a NaOH are needed. Molarity = moles of solute liter of solution no. of moles = molarity x volume = M.V Molarity moles gram NaOH Step 5 Calculation M.V = no. of moles 500 ml x 1 l x 3 mol3 NaOH = 1.5 mole 100 ml 1 l 1.5 moles NaOH x 40g NaoH = grams 1 mole NaOH Step 6 Answer Dissolve 60 grams of NaOH in a small amount of water and then add more water until the total volume is 500 mls.
Example Problem (2) Calculate the molarity of a solution that contains.036 grams HCl per 250 mls of solution. Step 1 - Given.036g HCl 250 mls of solution Step 2 - To find molarity M = moles liter g HC1 mole HCl molarity ml l Step 5 Calculations.036g HCl x 1000 ml x 1 mole HCl =.004 mole HCl 250 ml 1 l 36g HCl = 1l Step 6 Answer.004 Molar HCl preparation that uses the dilution method will use the dilution equation. It is: Where; M 1 V 1 = M 2 V 2 M 1 = molarity of the more concentrated solution V 1 = Volume in liters of M 1 used M 2 = Molarity of the solution to be prepared V 2 = Final total volume of M 2 (includes V 1 )
Example Problem (3) How many milliliters of 6 molar HCl must be used to prepare 500 mls of 1 molar solution. Step 1 To find V 1 = no. of mls of GM HCl Step 2 Given M 1 = 6M, V 2 = 500 ml, M 2 =1M M 1 V 1 =M 2 V 2 V 1 = M 2 V 2 solving for V 1 M 1 Step 5 Calculations V 1 = (1M) (500ml) (6M) V 1 = 88.3 mls of 6 M NaOH are needed to prepare 500 mls of 1 molar. Step 6 - Answer 83.3 ml of 6 M NaOH are needed to prepare 500 mls of 1 molar.
Example Problem (4) What volume of water must be added to 50 mls of 12 molar HCl to dilute it to 2 molar. Step 1 To find V 2 V 1 = no. of mls of water added. Step 2 Given V 1 = 50 ml, M 2 = 2M, M 1 = 12M M 1 V 1 =M 2 V 2 M1V V 2 = M 2 1 then no mls added = 300 ml 50 ml = 250 ml Step 6 Answer 250 mls of water must be added to 50 mls of 12M HCl to dilute it to 2 molar.
PROBLEMS 1. Define a solution and list its general properties 2. Describe how you would prepare the following: a. 600 mls of.2m NaOH b. 25 liters of.001m HNO 3 c..015 ml of.0000187 M H 2 SO 4 3. Calculate the molarities of the following mixtures a. 9.8g H 2 SO 4 250 mls of solution b. 53.6g NaOh 2500 ml of solution 4. What volume of 6M HN0 3 is needed to prepare 1500 mls of.15m. 5. What volume of H 2 O must be added to 50 mls of 15.8 molar NaOH to dilute it to 1.25 molar 2.