Solute and Solvent 7.1. Solutions. Examples of Solutions. Nature of Solutes in Solutions. Learning Check. Solution. Solutions

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1 Chapter 7 s 7.1 s Solute and Solvent s are homogeneous mixtures of two or more substances. consist of a solvent and one or more solutes. 1 2 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 4 Identify the solute in each of the following solutions. A. 2 g of sugar and 100 ml of water B ml of ethyl alcohol and 30.0 ml of methyl alcohol C ml of water and 1.50 g of NaCl Identify the solute in each of the following solutions. A. 2 g of sugar B ml of methyl alcohol C. 1.5 g of NaCl D. 200 ml of O 2 D. Air: 200 ml of O 2 and 800 ml of N 2 5 6

2 Water Formation of a Water is the most common solvent. is a polar molecule. forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule. Na + and Cl - ions on the surface of a NaCl crystal are attracted to polar water molecules. are hydrated in solution with many H 2 O molecules surrounding each ion. 7 8 Equations for Formation When NaCl(s) dissolves in water, the reaction can be written as NaCl(s) solid H 2 O Na + (aq) + Cl - (aq) 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. B. the Cl - ions are attracted to the 1) oxygen atom (δ - ) of water. 2) hydrogen atom (δ + )of water Like Dissolves Like 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 2) hydrogen atom (δ + )of water. Two substances form a solution when there is an attraction between the particles of the solute and solvent. when a polar solvent such as water dissolves polar solutes such as sugar, and ionic solutes such as NaCl. when a nonpolar solvent such as hexane (C 6 H 14 ) dissolves nonpolar solutes such as oil or grease

3 Water and a Polar Solute Like Dissolves Like Solvents Water (polar) CH 2 Cl 2 (nonpolar) Solutes Ni(NO 3 ) 2 (polar) 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, nonpolar 3) I 2 no, nonpolar 4) HCl yes, polar Most polar and ionic solutes dissolve in water because water is a polar solvent Chapter 7 s Solutes and Ionic Charge 7.2 Electrolytes and Nonelectrolytes In water, strong electrolytes produce ions and conduct an electric current. weak electrolytes produce a few ions. nonelectrolytes do not produce ions

4 Strong Electrolytes Strong electrolytes dissociate in water, producing positive and negative ions. conduct an electric current in water. in equations show the formation of ions in aqueous(aq) solutions. H 2 O 100% ions NaCl(s) Na + (aq) + Cl (aq) CaBr 2 (s) H 2 O Ca 2+ (aq) + 2Br (aq) Complete each for strong electrolytes in water. H 2 O A. CaCl 2 (s) 1) CaCl 2 (s) 2) Ca 2+ (aq) + Cl 2 (aq) 3) Ca 2+ (aq) + 2Cl (aq) H 2 O B. K 3 PO 4 (s) 1) 3K + (aq) + PO 3 4 (aq) 2) K 3 PO 4 (s) 3) K 3+ (aq) + P 3 (aq)+ O 4 (aq) Weak Electrolytes Complete each for strong electrolytes in water: H 2 O A. CaCl 2 (s) 3) Ca 2+ (aq) + 2Cl (aq) H 2 O B. K 3 PO 4 (s) 1) 3K + (aq) + PO 3 4 (aq) A weak electrolyte dissociates only slightly in water. in water forms a solution of a few ions and mostly undissociated molecules. HF(g) + H 2 O(l) H 3 O + (aq) + F - (aq) NH 3 (g) + H 2 O(l) NH 4+ (aq) + OH - (aq) Nonelectrolytes Nonelectrolytes dissolve as molecules in water. do not produce ions in water. do not conduct an electric current. Equivalents An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or -). 1 mole of Na + = 1 equivalent 1 mole of Cl = 1 equivalent 1 mole of Ca 2+ = 2 equivalents 1 mole of Fe 3+ = 3 equivalents 23 24

5 Electrolytes in Body Fluids Electrolytes in Body Fluids In replacement solutions for body fluids, the electrolytes are given in milliequivalents per liter (meq/l). Ringer s Na meq/l Cl 155 meq/l K + 4 meq/l Ca 2+ 4 meq/l The milliequivalents per liter of cations must equal the milliequivalents per liter of anions A. In 1 mole of Fe 3+, there are 1) 1 Eq. 2) 2 Eq. 3) 3 Eq. B. In 2.5 moles of SO 4 2, there are 1) 2.5 Eq. 2) 5.0 Eq. 3) 1.0 Eq. C. An IV bottle contains NaCl. If the Na + is 34 meq/l, the Cl is 1) 34 meq/l. 2) 0 meq/l. 3) 68 meq/l. A. 3) 3 Eq. B. 2) 5.0 Eq. 2.5 mole SO 2 4 x 2 Eq = 5.0 Eq 1 mole SO 2 4 C. 1) 34 meq/l Chapter 7 s Solubility 7.3 Solubility Solubility is the maximum amount of solute that dissolves in a specific amount of solvent. expressed as grams of solute in 100 grams of solvent, usually water. g of solute 100 g water 29 30

6 Unsaturated s Saturated s Unsaturated solutions contain less than the maximum amount of solute. can dissolve more solute. Saturated solutions contain the maximum amount of solute that can dissolve. have undissolved solute at the bottom of the container At 40 C, the solubility of KBr is 80 g/100 g of H 2 O. Identify the following solutions as either 1) saturated or 2) unsaturated. Explain. A. 60 g KBr added to 100 g of water at 40 C. B. 200 g KBr added to 200 g of water at 40 C. A. 2 Amount of 60 g KBr/100 g of water is less than the solubility of 80 g KBr/100 g of water. B. 1 In 100 g of water, 100 g of KBr exceeds the solubility of 80 g KBr /100 g of water at 40 C. C. 2 This is the same as 50 g KBr in 100 g of water, which is less than the solubility of 80 g KBr/ 100 g of water at 40 C. C. 25 g KBr added to 50 g of water at 40 C Effect of Temperature on Solubility Solubility depends on temperature. of most solids increases as temperature increases. of gases decreases as temperature increases. A. Why could a bottle of carbonated drink possibly burst (explode) when it is left out in the hot sun? B. Why do fish die in water that is too warm? 35 36

7 Solubility and Pressure A. The pressure in a bottle increases as the gas leaves solution when it becomes less soluble at higher temperatures. As pressure increases, the bottle could burst. B. Because O 2 gas is less soluble in warm water, fish cannot obtain the amount of O 2 required for their survival. Henry s law states the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid. at higher pressures, more gas molecules dissolve in the liquid Chapter 7 s Percent Concentration 7.4 Percent Concentration The concentration of a solution is the amount of solute dissolved in a specific amount of solution. amount of solute amount of solution Mass Percent Mass of Mass percent (% m/m) is the concentration by mass of solute in a solution. mass percent = g of solute x 100 g of solute + g of solvent amount in g of solute in 100 g of solution. mass percent = g of solute a 100 g of solution 8.00 g KCl Add water to give g of solution g of KCl solution 41 42

8 Calculating Mass Percent Guide to Calculating Concentrations The calculation of mass percent (% m/m) requires the grams of solute (g KCl) and grams of solution (g KCl solution). g of KCl = 8.00 g g of solvent (water) = g g of KCl solution = g 8.00 g KCl (solute) x 100 = 16.0% (m/m) g KCl solution A solution is prepared by mixing 15.0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. 1) 15.0% (m/m) Na 2 CO 3 2) 6.38% (m/m) Na 2 CO 3 3) 6.00% (m/m) Na 2 CO 3 3) 6.00% (m/m) Na 2 CO 3 STEP 1: mass solute = 15.0 g of Na 2 CO 3 mass solution = 15.0 g g = 250. g STEP 2: Use g solute/g solution ratio STEP 3: mass %(m/m) = g solute x 100 g solution STEP 4: Set up problem mass %(m/m) = 15.0 g Na 2 CO 3 x 100 = 6.00% Na 2 CO g solution Volume Percent Mass/Volume Percent The volume percent (% v/v) is percent volume (ml) of solute (liquid) to volume (ml) of solution. volume % (v/v) = ml of solute x 100 ml of solution solute (ml) in 100 ml of solution. volume % (v/v) = ml of solute 100 ml of solution The mass/volume percent (% m/v) is percent mass (g) of solute to volume (ml) of solution. mass/volume % (m/v) = g of solute x 100 ml of solution solute (g) in 100 ml of solution. mass/volume % (m/v) = g of solute 100 ml of solution 47 48

9 Percent Conversion Factors Two conversion factors can be written for each type of % value. Write two conversion factors for each solutions. A. 8.50% (m/m) NaOH B. 5.75% (v/v) ethanol C. 4.8% (m/v) HCl A g NaOH and 100 g solution 100 g solution 8.50 g NaOH B ml alcohol and 100 ml solution 100 ml solution 5.75 ml alcohol C. 4.8 g HCl and 100 ml HCl 100 ml solution 4.8 g HCl Using Percent Concentration (m/m) as Conversion Factors How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1: Given: 225 g solution; 10.0% (m/m) NaCl Need: g of NaCl STEP 2: g solution g NaCl STEP 3: Write the 10.0% (m/m) as conversion factors g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4: Set up using the factor that cancels g solution. 225 g solution x 10.0 g NaCl = 22.5 g NaCl 100 g solution How many grams of NaOH are needed to prepare 75.0 g of 14.0% (m/m) NaOH solution? 1) 10.5 g of NaOH 2) 75.0 g of NaOH 3) 536 g of NaOH 1) 10.5 g NaOH 75.0 g solution x 14.0 g NaOH = 10.5 g NaOH 100 g solution 14.0% (m/m) factor 53 54

10 How many milliliters of a 5.75% (v/v) ethanol solution can be prepared from 2.25 ml of ethanol? 1) 2.56 ml 2) 12.9 ml 3) 39.1 ml 3) 39.1 ml 2.25 ml ethanol x 100 ml solution 5.75 ml ethanol 5.75% (v/v) inverted = 39.1 ml of solution Using Percent Concentration (m/v) as Conversion Factors How many ml of a 4.20% (m/v) will contain 3.15 g of KCl? STEP 1: Given: 3.15 g of KCl(solute); 4.20% (m/v) KCl Need: ml of KCl solution STEP 2: Plan: g of KCl ml of KCl solution STEP 3: Write conversion factors g KCl and 100 ml solution 100 ml solution 4.20 g KCl STEP 4: Set up the problem 3.15 g KCl x 100 ml KCl solution = 75.0 ml of KCl 4.20 g KCl How many grams of NaOH are needed to prepare 125 ml of a 8.80% (m/v) NaOH solution? Chapter 7 s How many grams of NaOH are needed to prepare 125 ml of a 8.80% (m/v) NaOH solution? 125 ml solution x 8.80 g NaOH = 11.0 g of NaOH 100 ml solution 7.5 Molarity and Dilution 59 60

11 Molarity (M) Molarity (M) is a concentration term for solutions. gives the moles of solute in 1 L of solution. moles of solute liter of solution Preparing a 1.0 Molar A 1.00 M NaCl solution is prepared by weighing out 58.5 g of NaCl (1.00 mole) and adding water to make 1.00 liter of solution Calculation of Molarity What is the molarity of L of NaOH solution if it contains 6.00 g of NaOH? STEP 1: Given 6.00 g of NaOH in L of solution Need molarity (mole/l) STEP 2: Plan g NaOH mole NaOH molarity Calculation of Molarity (continued) STEP 3: Conversion factors 1 mole of NaOH = 40.0 g of NaOH 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4: Calculate molarity g NaOH x 1 mole NaOH = mole 40.0 g NaOH mole = mole = M NaOH L 1 L What is the molarity of 325 ml of a solution containing 46.8 g of NaHCO 3? 1) M 2) 1.44 M 3) 1.71 M 3) 1.71 M 46.8 g NaHCO 3 x 1 mole NaHCO 3 = mole NaHCO g NaHCO mole of NaHCO 3 = 1.71 M NaHCO L 65 66

12 What is the molarity of 225 ml of a KNO 3 solution containing 34.8 g of KNO 3? 1) M 2) 1.53 M 3) 15.5 M 2) 1.53 M 34.8 g KNO 3 x 1 mole KNO 3 = mole of KNO g KNO 3 M = mole = mole KNO 3 = 1.53 M L L In one setup: 34.8 g KNO 3 x 1 mole KNO 3 x 1 = 1.53 M g KNO L Molarity Conversion Factors Calculations Using Molarity The units of molarity are used as conversion factors in calculations with solutions. Molarity Equality 3.5 M HCl 1 L = 3.5 moles of HCl Written as Conversion Factors 3.5 moles HCl and 1 L 1 L 3.5 moles HCl How many grams of KCl are needed to prepare 125 ml of a M KCl solution? STEP 1: Given 125 ml (0.125 L) of M KCl Need g of KCl STEP 2: Plan L KCl moles KCl g KCl Calculations Using Molarity STEP 3: Conversion factors 1 mole of KCl = 74.6 g 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl 1 L KCl = mole of KCl 1 L and mole KCl mole KCl 1 L How many grams of AlCl 3 are needed to prepare 125 ml of a M solution? 1) 20.0 g of AlCl 3 2) 16.7 g of AlCl 3 3) 2.50 g of AlCl 3 STEP 4: Calculate grams L x mole KCl x 74.6 g KCl = 6.71 g of KCl 1 L 1 mole KCl 71 72

13 3) 2.50 g AlCl L x mole x g = 2.50 g of AlCl 3 1 L 1 mole How many milliliters of 2.00 M HNO 3 contain 24.0 g of HNO 3? 1) 12.0 ml 2) 83.3 ml 3) 190. ml g HNO 3 x 1 mole HNO 3 x 1000 ml 63.0 g HNO mole HNO 3 Molarity factor inverted Dilution In a dilution water is added. volume increases. concentration decreases. = 190. ml of HNO Comparing Initial and Diluted s Guide to Calculating Dilution Quantities In the initial and diluted solution, the moles of solute are the same. the concentrations and volumes are related by the following equations: For percent concentration: C 1 V 1 = C 2 V 2 initial diluted For molarity: M 1 V 1 = M 2 V 2 initial diluted 77 78

14 Dilution Calculations with Percent What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 ml of 14.0% (m/v) HCl solution? Prepare a table: C 1 = 14.0% (m/v) V 1 = 25.0 ml C 2 = 2.00% (m/v) V 2 =? Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 What is the percent (% m/v) of a solution prepared by diluting 10.0 ml of 9.00% NaOH to 60.0 ml? V 2 = V 1 C 1 = (25.0 ml)(14.0%) = 175 ml C % Dilution Calculations with Molarity What is the percent (% m/v) of a solution prepared by diluting 10.0 ml of 9.00% NaOH to 60.0 ml? Prepare a table: C 1 = 9.00 %(m/v) V 1 = 10.0 ml C 2 =? V 2 = 60.0 ml Solve dilution equation for unknown and enter values: C 1 V 1 = C 2 V 2 C 2 = C 1 V 1 = (10.0 ml)(9.00%) = 1.50% (m/v) V ml What is the molarity (M) of a solution prepared by diluting L of M HNO 3 to L? Prepare a table: M 1 = M V 1 = L M 2 =? V 2 = L Solve dilution equation for unknown and enter values: M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 = (0.600 M)(0.180 L) = M V L What is the final volume (ml) of 15.0 ml of a 1.80 M KOH diluted to give a M solution? 1) 27.0 ml 2) 60.0 ml 3) 90.0 ml What is the final volume (ml) of 15.0 ml of a 1.80 M KOH diluted to give a M solution? Prepare a table: M 1 = 1.80 M V 1 = 15.0 ml M 2 = 0.300M V 2 =? Solve dilution equation for V 2 and enter values: M 1 V 1 = M 2 V 2 V 2 = M 1 V 1 = (1.80 M)(15.0 ml) = 90.0 ml M M 83 84

15 Chapter 7 s Molarity in Chemical Reactions 7.6 s in Chemical Reactions In a chemical reaction, the volume and molarity of a solution are used to determine the moles of a reactant or product. molarity ( mole ) x volume (L) = moles 1 L if molarity (mole/l) and moles are given, the volume (L) can be determined. moles x 1 L = volume (L) moles Using Molarity of Reactants Using Molarity of Reactants (cont.) How many ml of 3.00 M HCl are needed to completely react with 4.85 g of CaCO 3? 2 HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 1: Given 3.00 M HCl; 4.85 g of CaCO 3 Need volume in ml STEP 2: Plan g CaCO 3 mole CaCO 3 mole HCl ml HCl 2HCl(aq) + CaCO 3 (s) CaCl 2 (aq) + CO 2 (g) + H 2 O(l) STEP 3: Equalities 1 mole of CaCO 3 = g of CaCO 3 1 mole of CaCO 3 = 2 moles of HCl 1000 ml of HCl = 3.00 moles of HCl STEP 4: Set Up 4.85 g CaCO 3 x 1 mole CaCO 3 x 2 mole HCl x 1000 ml HCl g CaCO 3 1 mole CaCO mole HCl = 32.3 ml of HCl required How many ml of a M Na 2 S solution are needed to completely react 18.5 ml of M NiCl 2 solution? NiCl 2 (aq) + Na 2 S(aq) 1) 4.16 ml 2) 6.24 ml 3) 27.0 ml NiS(s) + 2NaCl(aq) 3) 27.0 ml L x mole NiCl 2 x1 mole Na 2 S x 1000 ml 1 L 1 mole NiCl mole = 27.0 ml of Na 2 S solution 89 90

16 If 22.8 ml of M MgCl 2 is needed to completely react 15.0 ml of AgNO 3 solution, what is the molarity of the AgNO 3 solution? MgCl 2 (aq) + 2AgNO 3 (aq) 1) M 2) M 3) M 2AgCl(s) + Mg(NO 3 ) 2 (aq) 3) M AgNO L x mole MgCl 2 x2 mole AgNO 3 x 1 P 1 L 1 mole MgCl L = mole/l = M AgNO How many liters of H 2 gas at STP are produced when Zn reacts with 125 ml of 6.00 M HCl? Zn(s) + 2HCl(aq) 1) 4.20 L of H 2 2) 8.40 L of H 2 3) 16.8 L of H 2 ZnCl 2 (aq) + H 2 (g) 2) 8.40 L H 2 gas L x 6.00 mole HCl x 1 mole H 2 x 22.4 L 1 L 2 mole HCl 1 mole = 8.40 L of H 2 gas Chapter 7 s 7.7 Properties of s s s contain small particles (ions or molecules). are transparent. do not separate. cannot be filtered

17 Colloids Examples of Colloids Colloids have medium-size particles. cannot be filtered. can be separated by semipermeable membranes. Examples of colloids include fog whipped cream milk cheese blood plasma pearls Suspensions s, Colloids, and Suspensions Suspensions have very large particles. settle out. can be filtered. must be stirred to stay suspended. Examples include: blood platelets, muddy water, and calamine lotion A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a 1) solution. 2) colloid. 3) suspension. A mixture that has solute particles that do not settle out, but are too large to pass through a semipermeable membrane is called a 2) colloid

18 Osmosis Osmosis In osmosis, water (solvent) flows from the lower solute concentration into the higher solute concentration. the level of the solution with the higher solute concentration rises. the concentrations of the two solutions become equal with time. Suppose a semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semipermeable membrane 4% starch H 2 O 10% starch Water Flow Equalizes The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases. The 4% solution loses water and its volume decreases. Eventually, the water flow between the two becomes equal. 7% starch 7% starch H 2 O Osmotic Pressure Osmotic pressure is produced by the solute particles dissolved in a solution. equal to the pressure that would prevent the flow of additional water into the more concentrated solution. greater as the number of dissolved particles in the solution increases A semipermeable membrane separates a 10% sucrose solution A from a 5% sucrose solution B. If sucrose is a colloid, fill in the blanks in the statements below. 1. has the greater osmotic pressure. 2. Water initially flows from into. 3. The level of solution will be lower. A semipermeable membrane separates a 10% sucrose solution A from a 5% sucrose solution B. If sucrose is a colloid, fill in the blanks in the statements below. 1. A has the greater osmotic pressure. 2. Water initially flows from B into A. 3. The level of solution B will be lower

19 Osmotic Pressure of the Blood Isotonic s Red blood cells have cell walls that are semipermeable membranes. maintain an osmotic pressure that cannot change or damage occurs. must maintain an equal flow of water between the red blood cell and its surrounding environment. An isotonic solution exerts the same osmotic pressure as red blood cells. is known as a physiological solution. of 5.0% glucose or 0.90% NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells. H 2 O Hypotonic s Hypertonic s A hypotonic solution has a lower osmotic pressure than red blood cells. has a lower concentration than physiological solutions. causes water to flow into red blood cells. H 2 O causes hemolysis: RBCs swell and may burst. A hypertonic solution has a higher osmotic pressure than RBCs. has a higher concentration than physiological solutions. causes water to flow out of RBCs. cause crenation: RBCs shrink in size. H 2 O Dialysis In dialysis, solvent and small solute particles pass through an artificial membrane. large particles are retained inside. waste particles such as urea from blood are removed using hemodialysis (artificial kidney). Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A. 2% NaCl solution B. 1% glucose solution C. 0.5% NaCl solution D. 5% glucose solution

20 Indicate if each of the following solutions is 1) isotonic, 2) hypotonic, or 3) hypertonic. A. 3 2% NaCl solution B. 2 1% glucose solution C % NaCl solution D. 1 5% glucose solution When placed in each of the following, indicate if a red blood cell will 1) not change, 2) hemolyze, or 3) crenate. A. 5% glucose solution B. 1% glucose solution C. 0.5% NaCl solution D. 2% NaCl solution When placed in each of the following, indicate if a red blood cell will 1) not change, 2) hemolyze, or 3) crenate. A._1_ 5% glucose solution B._2_ 1% glucose solution C._2_ 0.5% NaCl solution D._3_ 2% NaCl solution Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag? A. 10% KCl solution B. 5% starch solution C. 5% NaCl and 5% starch solutions Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag? A. 10% KCl solution KCl ( K +, Cl ) B. 5% starch solution None; starch is retained. C. 5% NaCl and 5% starch solutions NaCl (Na +, Cl ), but starch is retained. 119

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