Chapter 15 Solutions
Key Terms for this Chapter Make sure you know the meaning of these: Solution Solute Solvent Aqueous solution Solubility Saturated Unsaturated Supersaturated Concentrated Dilute 15-2
Figure 15.1: Dissolving of solid sodium chloride. solute solution solvent 15-3
Figure 15.2: How Ionic substances dissolve: Polar water molecules interacting with positive and negative ions of a salt. 15-4
Dissolving Bonds Cations and anions are attracted to the polar water molecule. In a covalent bond, if the bond is polar, it will dissolve in water. Example: sugar in water. If the bond is NOT polar, the substance will not be attracted to water molecules and it will not dissolve in water. Example: oil in water. 15-5
Figure 15.6: An oil layer floating on water. 15-6
15.3 Factors Affecting the Rate of Dissolution If we dissolve sugar in water, what is the solute, and what is the solvent in this solution? What could we do to speed up the dissolving process? 15-7
Dissolving to Saturate or Supersaturate the Solution If we add sugar to a solution, we will eventually reach a point which sugar will no longer dissolve, and will begin to collect at the bottom. This is called saturation. Any point which before saturation, is known as an unsaturated solution. If we heat up the solution, we may be able to dissolve more solute than the solution can hold at that temperature, this is called supersaturation. Supersaturated solutions are very unstable, if even one crystal is added it could cause the immediate precipitation of the solid in the solution. 15-8
15.4 Solution Composition: Mass Percent One way to describe a solution s composition is through mass percent. % mass = mass of solute x 100 mass of solution 15-9
Example: Suppose a solution is prepared by dissolving 1.0 g of NaCl in 48 g of water. What percent of the solution, by mass, is the NaCl? 1.0 g solute x 100 = 2.0% NaCl 49 g solution 15-10
Practice Problems A solution is prepared by mixing 1.00 g of ethanol, C 2 H 5 OH, with 100.0 g of water. Calculate the mass percent of this solution. A laboratory assistant prepared a KCl solution for her class by dissolving 5.34 g of KCl in 152 g of water. What is the mass percent of the solution she prepared? What mass of each solute is present in 285 g of a solution that contains 5.00% by mass NaCl and 7.50% by mass by mass Na 2 CO 3? 15-11
Mass by Volume (v/v) If both the solvent and solute are liquids, then a solution can be made by measuring volumes. % volume = volume of solute x 100 volume of solution Example: 20 ml of rubbing alcohol is diluted with water to a total volume of 100 ml Calculate the % volume of alcohol. 15-12
Percent by mass & volume (m/v) If a solid is dissolved in liquid, you may know the mass of the solute and the volume of the liquid. % (m/v) = mass of solute x 100 volume of solution Example: A solution contains 7 grams of NaCl in 100 ml of solution. Calculate the percent by mass/volume 15-13
15.5 Molarity We define the concentration of a solution as the amount of solute in a given volume of solution. The most common way to express concentration is in molarity (M). Molarity is the number of moles of solute per volume of solution in liters. 15-14
Calculating Ion Concentration from Molarity When we look at a solution, we must understand that if 1.0 mol NaCl is dissolved in enough water to make 1.0 L of solution, there is not 1.0 mol of NaCl running around in the solution. When the NaCl is in the solution, it dissolves and become ions, Na + and Cl -. This means that if the NaCl dissolves completely, there is 1.0 mol Na + ions and 1.0 mol Cl - ions. NaCl Na + + Cl - 15-15
Example: For the following reaction, calculate the number of all of the ions in 0.50 M Co(NO 3 ) 2 Co(NO 3 ) 2(s) H Co 2+ (aq) + 2NO - 2 O (l) 3 (aq) When solid Co(NO 3 ) 2 dissolves, it produces the ions above, which we can write in terms of moles. 1mol Co(NO 3 ) 2(s) 1mol Co 2+ (aq) + 2mol NO - H 2 O (l) 3 In a solution with 0.50 M Co(NO 3 ) 2 there is 0.50 M Co 2+ and (2x0.50) M NO 3- or 1.0M NO - 3 15-16
Practice Problems Give the concentration of the ions in each of the following solutions: a. 1 M FeCl 3 b. 0.10 M Na 2 CO 3 c. 0.010 M Al 2 (SO 4 ) 3 15-17
Making a Stock Solution Now that you know how to calculate molarity, let s use this information to make a standard solution. A standard solution is a solution whose concentration is known. To make a standard solution, the mass of solute is calculated and weighed out. The solid is then added to the volumetric flask. Next solvent is added about halfway, the lid is placed on and the you swirl to dissolve the liquid. Finally the remainder of the solvent is added to the line, carefully read the bottom of the meniscus to the line. 15-18
Figure 15.7: Steps involved in the preparation of a standard aqueous solution. 15-19
15.6 Dilution Dilution: The process of adding more solvent to a solution. When we dilute a solution, we are only adding water to the solution. This means that the amount of solute in the solution has not changed. Moles of solute before dilution = moles of solute after dilution If the number of moles stays the same, but the volume of solution increases.what happens to the molarity? 15-20
Dilution Molarity We already know that if we know the volume of the solution and the molarity of the solution, we can calculate the moles of solute. Since the number of moles of solute do not change before or after the dilution, we can use this information in an equation to calculate the needed volumes and molarities. M 1 x V 1 = moles of solute = M 2 x V 2 Where M 1 V 1 is molarity and volume before the dilution, and M 2 V 2 is molarity and volume after the dilution. 15-21
Example: Suppose we want 500. ml of 1.00 M acetic acid from a 17.5 M stock solution of acetic acid. What volume of stock solution do we need to dilute? What information do we know? M 1 = 17.5 M M 2 = 1.00 M V 2 = 500. ml So using M 1 V 1 =M 2 V 2 what are we trying to solve for? Rearrange the equation and solve. When you have your answer, check to make sure that both solutions have equal amounts of moles. 15-22
Practice Problems What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a 0.10 M H 2 SO 4 solution? If I add 340mL of a 0.5M NaBr solution, what will the concentration be if I add 560mL more water? If I dilute 250mL of 0.10M lithium acetate solution to a volume of 750mL, what is the concentration? 15-23
How to Properly Dilute an Acidic Solution 1. Always add acid to the water NEVER the reverse. 2. If making our acetic acid solution, we calculated that we need 28.6 ml of 17.5 M acetic acid. 3. Fill the volumetric flask up partially with water. 4. Add the 28.6 ml of 17.5M acetic acid. 5. Add more water, swirl, more water, swirl, until solution is up to the line of the necessary volume, remember read the meniscus. 15-24
15.7 Stoichiometry of Solution Reactions Remember back to Ch. 9 when we were calculating reactants and products in chemical reactions. Solve this: What mass of O 2 is required to react with 3.6 grams of SO 2 in the following reaction? SO 2(g) + O 2(g) SO 3(g) 15-25
Stoichiometry in Solutions The steps we will take are as follows: 1. Write the balanced equation for the reaction. If the reaction involves ions, write the net ionic equation. 2. Calculate the moles of reactants. 3. Determine which reactant is limiting. 4. Calculate the moles of other reactants or products as required. 5. Convert to grams or other units, if required. 15-26
Example: Calculate the mass of solid NaCl that must be added to 1.50 L of a 0.100 M AgNO 3 solution to precipitate all of the Ag + ions in the form of AgCl. Calculate the mass of AgCl formed. Step 1: Write the balanced net ionic equation. AgNO 3(aq) + NaCl (aq) NaNO 3(aq) + AgCl (s) Ag + (aq) + Cl - (aq) AgCl (s) Step 2: Calculate the moles of reactants. 0.100 M AgNO 3 x 1 mol Ag + = 0.100 M Ag + 1 mol AgNO 3 1.50 L x 0.100 mol Ag + = mol Ag + 1 L 15-27
Step 3: Determine which reactant is limiting. In this situation we only want enough Cl - to react with the Ag + present. We want to precipitate all of the Ag +, thus the amount of Ag + present determines the amount of Cl - present. Step 4: Determine the moles of Cl - required. We have 0.150 mol of Ag + use a mole ratio to determine the amount of Cl -. 0.150 mol Ag + x 1 mol Cl - = mol Cl - 1 mol Ag + 15-28
Step 5: Calculate the mass of AgCl formed, and the mass of NaCl required. If we know there are 0.150 mol Cl - formed, we can calculate the moles of AgCl formed 0.150 mol Cl - x 1 mol AgCl = 0.150 mol AgCl 1 mol Cl - 0.150 mol AgCl x 143.3 g AgCl = 1 mol AgCl Now you calculate the mass of NaCl required. 15-29
Practice Problems When aqueous solutions of Na 2 SO 4 and Pb(NO 3 ) 2 are mixed, PbSO 4 precipitates. Calculate the mass of PbSO 4 formed when 1.25 L of 0.0500 M Pb(NO 3 ) 2 and 2.00 L of 0.0250 M Na 2 SO 4 are mixed. HINT: Calculate the moles of Pb 2+ and SO 4 2- in the mixed solution, decide which ion is limiting, and calculate the moles of PbSO 4 formed. 15-30