9.1 Water. Chapter 9 Solutions. Water. Water in Foods

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Chapter 9 s 9.1 Water 9.1 Properties of Water 9.2 s 9.3 Electrolytes and Nonelectrolytes 9.6 Percent Concentration 9.7 Molarity Water is the most common solvent. The water molecule is polar.

1 Chapter 9 s 9.1 Water 9.1 Properties of Water 9.2 s 9.3 Electrolytes and Nonelectrolytes 9.6 Percent Concentration 9.7 Molarity Water is the most common solvent. The water molecule is polar. Hydrogen bonds form between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. 1 2 Water Water in Foods Water for the body is obtained from fluids as well as foods. Some foods have a high percentage of water

surface, which pulls them closer. At the surface behave like a thin, elastic membrane, or skin.

s Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes.

2 Surface Tension 9.2 s: Solute and Solvent Water molecules At the surface form hydrogen bonds with molecules on or below the surface, which pulls them closer. At the surface behave like a thin, elastic membrane, or skin. Cannot hydrogen bond when compounds called surfactants are added. s Are homogeneous mixtures of two or more substances. Consist of a solvent and one or more solutes. 5 6 Nature of Solutes in s Examples of s Solutes Spread evenly throughout the solution. Cannot be separated by filtration. Can be separated by evaporation. Are not visible, but can give a color to the solution. The solute and solvent in a solution can be a solid, liquid, and/or a gas

3 Identify the solute and the solvent in each. A. brass: 20 g zinc + 50 g copper solute = 1) zinc 2) copper solvent = 1) zinc 2) copper B. 100 g H 2 O + 5 g KCl solute = 1) KCl 2) H 2 O solvent = 1) KCl 2) H 2 O Identify the solute and the solvent in each. A. brass: 20 g zinc + 50 g copper solute = 1) zinc solvent = 2) copper B. 100 g H 2 O + 5 g KCl solute = 1) KCl solvent = 2) H 2 O 9 10 Identify the solute in each of the following solutions: A. 2 g sugar (1) and 100 ml water (2) B ml of ethyl alcohol (1) and 30.0 ml of methyl alcohol (2) C ml water (1) and 1.50 g NaCl (2) D. Air: 200 ml O 2 (1) and 800 ml N 2 (2) Identify the solute in each of the following solutions: A. 2 g sugar (1) B ml of methyl alcohol (2) C. 1.5 g NaCl (2) D. 200 ml O 2 (1)

Like Dissolves Like Examples of Like Dissolves Like A solution forms when there is an attraction between the particles of the solute and solvent.

4 Like Dissolves Like Examples of Like Dissolves Like A solution forms when there is an attraction between the particles of the solute and solvent. A polar solvent such as water dissolves polar solutes such as sugar and ionic solutes such as NaCl. A nonpolar solvent such as hexane (C 6 H 14 ) dissolves nonpolar solutes such as oil or grease. Solvents Solutes Water (polar) Ni(NO 3 ) 2 (ionic) CH 2 Cl 2 (nonpolar) I 2 (nonpolar) Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 2) gasoline (nonpolar) 3) I 2 4) HCl Which of the following solutes will dissolve in water? Why? 1) Na 2 SO 4 Yes, ionic 2) gasoline No, nonnpolar 3) I 2 No, nonpolar 4) HCl Yes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent

Formation of a Na + and Cl - ions on the surface of a NaCl crystal are attracted to polar water molecules. In solution, the ions are hydrated as several H 2 O molecules surround each.

5 Formation of a Na + and Cl - ions on the surface of a NaCl crystal are attracted to polar water molecules. In solution, the ions are hydrated as several H 2 O molecules surround each. Equations for Formation When NaCl(s) dissolves in water, the reaction can be written as NaCl(s) H 2 O Na + (aq) + Cl (aq) solid separation of ions Solid LiCl is added to water. It dissolves because A. The Li + ions are attracted to the 1) oxygen atom (δ ) of water. 2) hydrogen atom (δ + ) of water. Solid LiCl is added to water. It dissolves because A. The Li + ions are attracted to the 1) oxygen atom (δ ) of water. B. The Cl - ions are attracted to the 1) oxygen atom (δ ) of water. 2) hydrogen atom (δ + ) of water. B. The Cl - ions are attracted to the 2) hydrogen atom (δ + ) of water

Heating a hydrate releases the water of hydration to give the anhydrate salt.

6 Hydrates Examples of Hydrates Hydrates are solid compounds that contain water molecules as part of the crystal structure. Heating a hydrate releases the water of hydration to give the anhydrate salt Write the equation for the dehydration of AlCl 3 6H 2 O. Write the equation for the dehydration of AlCl 3 6H 2 O. AlCl 3 6H 2 O AlCl 3 + 6H 2 O

9.3 Electrolytes Strong Electrolytes Electrolytes Produce positive (+) and negative (-) ions when they dissolve in water. In water conduct an electric current.

7 9.3 Electrolytes Strong Electrolytes Electrolytes Produce positive (+) and negative (-) ions when they dissolve in water. In water conduct an electric current. Strong electrolytes ionize 100% in solution. Equations for the dissociation of strong electrolytes show the formation of ions in aqueous (aq) solutions. H 2 O 100% ions NaCl(s) Na + (aq) + Cl - (aq) H 2 O CaBr 2 (s) Ca 2+ (aq) + 2Br - (aq) Complete each of the following dissociation equations for strong electrolytes dissolved in water: H 2 O A. CaCl 2 (s) 1) CaCl 2 2) Ca 2+ + Cl - 2 3) Ca Cl - H 2 O B. K 3 PO 4 (s) 1) 3K + + PO 3-4 Complete each of the following dissociation equations for strong electrolytes dissolved in water: H 2 O A. 3) CaCl 2 (s) Ca Cl - H 2 O B. 1) K 3 PO 4 (s) 3K + + PO 4 3-2) K 3 PO 4 3) K 3+ + P 3- + O

Weak Electrolytes A weak electrolyte Dissolves mostly as molecules in solution. Produces only a few ions in aqueous solutions. Has an equilibrium that favors the reactants.

8 Weak Electrolytes A weak electrolyte Dissolves mostly as molecules in solution. Produces only a few ions in aqueous solutions. Has an equilibrium that favors the reactants. HF + H 2 O H 3 O + (aq) + F - (aq) Nonelectrolytes Nonelectrolytes Form only molecules in water. Do not produce ions in water. Do not conduct an electric current. NH 3 + H 2 O NH 4+ (aq) + OH - (aq) Percent Concentration Mass Percent The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution The percent concentration describes the amount of solute that is dissolved in 100 parts of solution. amount of solute 100 parts solution The mass percent (%m/m) Concentration is the percent by mass of solute in a solution. mass percent = g of solute x 100% g of solution Is the g of solute in 100 g of solution. mass percent = g of solute 100 g of solution

Mass of grams of solute + grams of solvent 50.0 g KCl solution Calculating Mass Percent Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution).

9 Mass of grams of solute + grams of solvent 50.0 g KCl solution Calculating Mass Percent Mass percent (%m/m) is calculated from the grams of solute (g KCl) and the grams of solution (g KCl solution). g of KCl = 8.0 g g of solvent (water) = 42.0 g g of KCl solution = 50.0 g 8.0 g KCl (solute) x 100 = 16% (m/m) KCl 50.0 g KCl solution A solution is prepared by mixing 15 g Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (%m/m) of the solution. 1) 15% (m/m) Na 2 CO 3 3) 6.0% (m/m) Na 2 CO 3 mass solute = 15 g Na 2 CO 3 mass solution = 15 g g = 250 g 2) 6.4% (m/m) Na 2 CO 3 3) 6.0% (m/m) Na 2 CO 3 mass %(m/m) = 15 g Na 2 CO 3 x g solution = 6.0% Na 2 CO 3 solution

mass/volume % = g of solute 100 ml of solution Preparing a with a Mass/Volume % Concentration A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to

10 Mass/Volume Percent The mass/volume percent (%m/v) Concentration is the ratio of the mass in grams (g) of solute in a volume (ml) of solution. mass/volume % = g of solute x 100% ml of solution Is the g of solute in 100 ml of solution. mass/volume % = g of solute 100 ml of solution Preparing a with a Mass/Volume % Concentration A percent mass/volume solution is prepared by weighing out the grams of solute (g) and adding water to give the final volume of the solution Calculation of Mass/Volume Percent Mass/volume percent (%m/v) is calculated from the grams of solute (g KCl) and the volume of solution (ml KCl solution). g of KI = 5.0 g KI ml of KI solution = ml 5.0 g KI (solute) x 100 = 2.0%(m/v) KI ml KI solution A 500. ml samples of an IV glucose solution contains 25 g glucose (C 6 H 12 O 6 ) in water. What is the mass/volume % (%m/v) of glucose of the IV solution? 1) 5.0% 2) 20.% 3) 50.%

Volume Percent 1) 5.0% Mass/volume %(m/v) = 25 g glucose x 100 500. ml solution The volume percent (%v/v) Concentration is the percent volume (ml) of solute (liquid) to volume (ml) of solution.

11 Volume Percent 1) 5.0% Mass/volume %(m/v) = 25 g glucose x ml solution The volume percent (%v/v) Concentration is the percent volume (ml) of solute (liquid) to volume (ml) of solution. volume % (v/v) = ml of solute x 100% ml of solution = 5.0 % (m/v) glucose solution Is the ml of solute in 100 ml of solution. volume % (v/v) = ml of solute 100 ml of solution Percent Conversion Factors Two conversion factors can be written for any type of % value. Write two conversion factors for each solutions: A. 8%(m/v) NaOH B. 12%(v/v) ethyl alcohol

12 Using Percent Factors A. 8%(m/v) NaOH 8 g NaOH and 100 ml solution 100 ml solution 8 g NaOH B. 12%(v/v) ethyl alcohol 12 ml alcohol and 100 ml solution How many grams of NaCl are needed to prepare 250 g of a 10.0% (m/m) NaCl solution? 1. Write the 10.0 % (m/m) as conversion factors g NaCl and 100 g solution 100 g solution 10.0 g NaCl 2. Use the factor that cancels given (g solution). 100 ml solution 12 ml alcohol 250 g solution x 10.0 g NaCl = 25 g NaCl 100 g solution How many grams of NaOH are needed to prepare 2.0 L of a 12%(m/v) NaOH solution? 1) 24 g NaOH 2) 240 g NaOH 3) 2400 g NaOH 2) 240 g NaOH 2.0 L x 1000 ml = 2000 ml 1 L 2000 ml x 12 g NaOH = 240 g NaOH 100 ml 12 % (m/v) factor

13 How many milliliters of 5% (m/v) glucose solution are given if a patient receives 150 g of glucose? 1) 30 ml 2) 3000 ml 3) 7500 ml 2) 3000 ml 150 g glucose x 100 ml = 3000 ml 5 g glucose 5% m/v factor (inverted) Molarity (M) Preparing a 1.0 Molar Molarity is a concentration unit for the moles of solute in the liters (L) of solution. A 1.0 M NaCl solution is prepared by weighing out 58.5 g NaCl ( 1.0 mole) and adding water to make 1.0 liter of solution. Molarity (M) = moles of solute = moles liter of solution L Examples: 2.0 M HCl = 2.0 moles HCl 1 L 6.0 M HCl = 6.0 moles HCl 1 L

14 Calculation of Molarity What is the molarity of a NaOH solution prepared by adding 4.0 g of solid NaOH to water to make 0.50 L of solution? 1. Determine the moles of solute. 4.0 g NaOH x 1 mole NaOH = 0.10 mole 40.0 g NaOH 2. Calculate molarity mole = 0.20 mole = 0.20 M NaOH 0.50 L 1 L Calculate the molarity of an NaHCO 3 solution prepared by dissolving 36 g of solid NaHCO 3 in water to give a solution volume of 240 ml. 1) 0.43 M 2) 1.8 M 3) 15 M ) 1.8 M 36 g x 1 mole NaHCO 3 = 0.43 mole NaHCO 3 84 g 0.43 mole NaHCO 3 = 1.8 M NaHCO L A glucose solution with a volume of 2.0 L contains 72 g glucose (C 6 H 12 O 6 ). If glucose has a molar mass of 180. g/mole, what is the molarity of the glucose solution? 1) 0.20 M 2) 5.0 M 3) 36 M

1) 0.20 M 72 g x 1 mole x 1 = 0.20 moles 180. g 2.0 L 1 L Molarity Conversion Factors The units in molarity can be used to write conversion factors. = 0.20 M 57 58 Stomach acid is 0.10 M HCl solution.

15 1) 0.20 M 72 g x 1 mole x 1 = 0.20 moles 180. g 2.0 L 1 L Molarity Conversion Factors The units in molarity can be used to write conversion factors. = 0.20 M Stomach acid is 0.10 M HCl solution. How many moles of HCl are present in 1500 ml of stomach acid? 1) 15 moles HCl 2) 1.5 moles HCl 3) 0.15 mole HCl 3) 0.15 mole HCl 1500 ml x 1 L = 1.5 L 1000 ml 1.5 L x 0.10 mole HCl = 0.15 mole HCl 1 L Molarity factor

16 Calculate the grams of KCl that must be dissolved in water to prepare 0.25 L of a 2.0 M KCl solution. 1) 150 g KCl 2) 37 g KCl 3) 19 g KCl 3) 37 g KCl Determine the number of moles of KCl L x 2.0 mole KCl = 0.50 moles KCl 1 L Convert the moles to grams of KCl moles KCl x 74.6 g KCl = 37 g KCl 1 mole KCl molar mass of KCl How many milliliters of 6.0 M HNO 3 contain 0.15 mole of HNO 3? 1) 25 ml 2) 90 ml 3) 400 ml 1) 25 ml 0.15 mole HNO 3 x 1 L x 1000 ml 6.0 moles HNO 3 1 L Molarity factor inverted = 25 ml HNO

Chemistry 51 Chapter 8 TYPES OF SOLUTIONS. Some Examples of Solutions. Type Example Solute Solvent Gas in gas Air Oxygen (gas) Nitrogen (gas)

Chemistry 51 Chapter 8 TYPES OF SOLUTIONS. Some Examples of Solutions. Type Example Solute Solvent Gas in gas Air Oxygen (gas) Nitrogen (gas) TYPES OF SOLUTIONS A solution is a homogeneous mixture of two substances: a solute and a solvent. Solute: substance being dissolved; present in lesser amount. Solvent: substance doing the dissolving; present