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Chapter 8 Solutions Concentration 8.4 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 1 Mass Percent Concentration 2 Mass of Solution Mass percent (%m/m) concentration is the • Percent by mass of solute in a solution. mass percent (%m/m) = g of solute x 100 8.00 g KCl Add water to give 50.00 g solution g of solute + g of solvent • Amount in g of solute in 100 g of solution. mass percent = g of solute 50.00 g KCl solution x 100 100 g of solution 3 4 1 Calculating Mass Percent Learning Check 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) = 42.00 g g of KCl solution = 50.00 g A solution is prepared by mixing 15.0 g Na2CO3 and 235 g of H2O. Calculate the mass percent (%m/m) of the solution. 8.00 g KCl (solute) 50.00 g KCl solution 1) 15.0% (m/m) Na2CO3 2) 6.38% (m/m) Na2CO3 3) 6.00% (m/m) Na2CO3 x 100 = 16.0% (m/m) 5 Solution 6 Volume Percent 3) 6.00% (m/m) Na2CO3 The volume percent (%v/v) is STEP 1 mass solute § Percent volume (mL) of solute (liquid) to volume (mL) of solution. = 15.0 g Na2CO3 mass solution = 15.0 g + 235 g = 250. g § Volume % (v/v) = 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 Na2CO3 x 100 = 6.00% Na2CO3 250. g solution 7 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 8 2 Mass/Volume Percent Percent Conversion Factors § Two conversion factors can be written for each type of % value. 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 TABLE 8.9 § Solute (g) in 100 mL of solution. mass/volume % (m/v) = g of solute 100 mL of solution 5% (m/v) glucose There are 5 g of glucose in 100 mL of solution. 5 g glucose 100 mL solution and 100 mL solution 5 g glucose 9 10 Learning Check Solution Write two conversion factors for each solution: A. 8.50 g NaOH A. 8.50%(m/m) NaOH and 100 g solution B. 5.75 mL alcohol and B. 5.75%(v/v) ethanol 100 mL solution C. 4.8 g HCl C. 4.8 %(m/v) HCl 100 mL solution 11 100 g solution 8.50 g NaOH 100 mL solution 5.75 mL alcohol and 100 mL HCl 4.8 g HCl 12 3 Using Percent Concentration (m/ m) Factors Learning Check 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. 10.0 g NaCl and 100 g solution 100 g solution 10.0 g NaCl STEP 4 Set up to cancel 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 NaOH 2) 75.0 g NaOH 3) 536 g NaOH 13 14 Solution Learning Check 1) 10.5 g NaOH How many milliliters of a 5.75 % (v/v) ethanol solution can be prepared from 2.25 mL ethanol? 75.0 g solution x 14.0 g NaOH = 10.5 g NaOH 100 g solution 14.0 % (m/m) factor 1) 2.56 mL 2) 12.9 mL 3) 39.1 mL 15 16 4 Using Percent Concentration(m/v) Factors Solution How many mL of a 4.20%(m/v) will contain 3.15 g KCl? 3) 39.1 mL STEP 1 Given: 3.15 g KCl(solute); 4.20% (m/v) KCl Need: mL of KCl solution STEP 2 Plan: g KCl mL KCl solution STEP 3 Write conversion factors. 4.20 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 KCl 4.20 g KCl 2.25 mL ethanol x 100 mL solution 5.75 mL ethanol 5.75 %(v/v) inverted = 39.1 mL solution 17 18 Learning Check Solution How many grams of NaOH are needed to prepare 125 mL of a 8.80%(m/v) NaOH solution? 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 = 100 mL solution 19 11.0 g NaOH 20 5 Chapter 8 Solutions Molarity (M) 8.5 Molarity and Dilution Molarity (M) § Is a concentration term for solutions. § Gives the moles of solute in 1 L solution. § = moles of solute liter of solution 21 22 Preparing a 1.0 Molar Solution Calculation of Molarity A 1.00 M NaCl solution is prepared § By weighing out 58.5 g NaCl (1.00 mole) and § Adding water to make 1.00 liter of solution. What is the molarity of 0.500 L NaOH solution if it contains 6.00 g NaOH? STEP 1 Given 6.00 g NaOH in 0.500 L solution Need molarity (mole/L) STEP 2 Plan g NaOH 23 mole NaOH molarity 24 6 Calculation of Molarity (cont.) Learning Check What is the molarity of 325 mL of a solution containing 46.8 g of NaHCO3? STEP 3 Conversion factors 1 mole NaOH = 40.0 g 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Calculate molarity. 6.00 g NaOH x 1 mole NaOH = 0.150 mole 40.0 g NaOH 0.150 mole = 0.300 mole = 0.300 M NaOH 0.500 L 1L 1) 0.557 M 2) 1.44 M 3) 1.71 M 25 Solution 26 Learning Check 3) 1.71 M 46.8 g NaHCO3 x 1 mole NaHCO3 = 0.557 mole NaHCO3 84.0 g NaHCO3 What is the molarity of 225 mL of a KNO3 solution containing 34.8 g KNO3? 1) 0.344 M 2) 1.53 M 3) 15.5 M 0.557 mole NaHCO3 = 1.71 M NaHCO3 0.325 L 27 28 7 Solution Molarity Conversion Factors 2) 1.53 M 34.8 g KNO3 x 1 mole KNO3 101.1 g KNO3 The units of molarity are used to write conversion factors for calculations with solutions. = 0.344 mole KNO3 TABLE 8.10 M = mole = 0.344 mole KNO3 = 1.53 M L 0.225 L In one setup 34.8 g KNO3 x 1 mole KNO3 x 1 = 1.53 M 101.1 g KNO3 0.225 L 29 30 Calculations Using Molarity Calculations Using Molarity How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 3 Conversion factors 1 mole KCl = 74.6 g 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl STEP 1 Given 125 mL (0.125 L) of 0.720 M KCl Need Grams of KCl STEP 2 Plan L KCl moles KCl 1 L KCl = 0.720 mole KCl 1L and 0.720 mole KCl 0.720 mole KCl 1L g KCl STEP 4 Calculate g KCl 0.125 L x 0.720 mole KCl x 74.6 g KCl 1L 1 mole KCl 31 = 6.71 g KCl 32 8 Learning Check Solution 3) How many grams of AlCl3 are needed to prepare 125 mL of a 0.150 M solution? 1) 20.0 g AlCl3 2) 16.7g AlCl3 3) 2.50 g AlCl3 2.50 g AlCl3 0.125 L x 0.150 mole x 133.5 g = 2.50 g AlCl3 1L 1 mole 33 Learning Check 34 Solution How many milliliters of 2.00 M HNO3 contain 24.0 g HNO3? 24.0 g HNO3 x 1 mole HNO3 x 63.0 g HNO3 1000 mL = 2.00 moles HNO3 Molarity factor inverted 1) 12.0 mL 2) 83.3 mL 3) 190. mL = 190. mL HNO3 35 36 9 Dilution Initial and Diluted Solutions In a dilution, 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 C1V1 = C2V2 § Water is added. § Volume increases. § Concentration decreases. initial diluted For molarity M1V1 = M2V2 initial diluted 37 38 Dilution Calculations with Percent Learning Check What volume of a 2.00 %(m/v) HCl solution can be What is the percent (%m/v) of a solution prepared prepared by diluting 25.0 mL of 14.0%(m/v) HCl solution? by diluting 10.0 mL of 9.00% NaOH to 60.0 mL? Prepare a table: C1= 14.0 %(m/v) V1 = 25.0 mL C2= 2.00%(m/v) V2 = ? Solve dilution equation for unknown and enter values: C1V1 = C2V2 V2 = V1C1 C2 = (25.0 mL)(14.0%) = 175 mL 2.00% 39 40 10 Solution Dilution Calculations 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: C1= 9.00 %(m/v) V1 = 10.0 mL C 2= ? V2 = 60.0 mL Solve dilution equation for unknown and enter values: C1V1 = C2V2 What is the molarity (M) of a solution prepared by diluting 0.180L of 0.600 M HNO3 to 0.540 L? Prepare a table: M1= 0.600 M V1 = 0.180 L M2 = ? V2 = 0.540 L Solve dilution equation for unknown and enter values: M1V1 = M2V2 C2 = C1 V1 = (10.0 mL)(9.00%) = 1.50 %(m/v) V2 60.0 mL M2 = M1V1 V2 = (0.600 M)(0.180 L) = 0.200 M 0.540 L 41 42 Learning Check Solution What is the final volume (mL) of 15.0 mL of a 1.80 M What is the final volume (mL) of 15.0 mL of a 1.80 M KOH diluted to give a 0.300 M solution? Prepare a table: M1= 1.80 M V1 = 15.0 mL M2= 0.300M V2 = ? Solve dilution equation for V2 and enter values: M1V1 = M2V2 KOH diluted to give a 0.300 M solution? 1) 27.0 mL 2) 60.0 mL 3) 90.0 mL V2 43 = M1V1 M2 = (1.80 M)(15.0 mL) = 90.0 mL 0.300 M 44 11 Molarity in Chemical Reactions Using Molarity of Reactants In a chemical reaction, § The volume and molarity of a solution are used to determine the moles of a reactant or product. How many mL of 3.00 M HCl are needed to completely react with 4.85 g CaCO3? 2HCl(aq) + CaCO3(s) molarity ( mole ) x volume (L) = moles 1L § If molarity (mole/L) and moles are given, the volume (L) can be determined moles x 1L = volume (L) moles CaCl2(aq) + CO2(g) + H2O(l) STEP 1 Given 3.00 M HCl; 4.85 g CaCO3 Need volume in mL STEP 2 Plan g CaCO3 mole CaCO3 mole HCl mL HCl 45 Using Molarity of Reactants (cont.) 2HCl(aq) + CaCO3(s) Learning Check CaCl2(aq) + CO2(g) + H2O(l) STEP 3 Equalitites How many mL of a 0.150 M Na2S solution are needed to completely react 18.5 mL of 0.225 M NiCl2 solution? 1 mole CaCO3 = 100.1 g; 1 mole CaCO3 = 2 mole HCl 1000 mL HCl = 3.00 mole HCl STEP 4 Set Up 4.85 g CaCO3 x 1 mole CaCO3 x 2 mole HCl x 1000 mL HCl 100.1 g CaCO3 46 1 mole CaCO3 3.00 mole HCl NiCl2(aq) + Na2S(aq) NiS(s) + 2NaCl(aq) 1) 4.16 mL 2) 6.24 mL 3) 27.8 mL = 32.3 mL HCl required 47 48 12 Solution Learning Check If 22.8 mL of 0.100 M MgCl2 is needed to completely react 15.0 mL of AgNO3 solution, what is the molarity of the AgNO3 solution? 3) 27.8 mL 0.0185 L x 0.225 mole NiCl2 x 1 mole Na2S x 1000 mL 1L 1 mole NiCl2 0.150 mole Na2S MgCl2(aq) + 2AgNO3(aq) = 27.8 mL Na2S solution 2AgCl(s) + Mg(NO3)2(aq) 1) 0.0760 M 2) 0.152 M 3) 0.304 M 49 50 Solution Learning Check 3) 0.304 M AgNO3 How many liters of H2 gas at STP are produced when Zn react with 125 mL of 6.00 M HCl? Zn(s) + 2HCl(aq) ZnCl2 (aq) + H2(g) 0.0228 L x 0.100 mole MgCl2 x 2 moles AgNO3 x 1L 1 mole MgCl2 1 0.0150 L 1) 4.20 L H2 2) 8.40 L H2 3) 16.8 L H2 = 0.304 mole/L = 0.304 M AgNO3 51 52 13 Solution Chapter 8 2) 8.40 L H2 gas Solutions 8.6 Properties of Solutions 0.125 L x 6.00 moles HCl x 1 mole H2 x 22.4 L 1L 2 moles HCl 1 mole H2 = 8.40 L H2 gas 53 54 Solutions Colloids Solutions Colloids § Contain small particles (ions or molecules). § Have medium size particles. § Are transparent. § Cannot be filtered. § Do not separate. § Can be separated by semipermeable membranes. § Cannot be filtered. § Scatter light (Tyndall effect). § Do not scatter light. 55 56 14 Examples of Colloids Suspensions Suspensions TABLE 8.11 § Have very large particles. § Settle out. § Can be filtered. § Must be stirred to stay suspended. § Examples include blood platelets, muddy water, and Calamine lotion. 57 Solutions, Colloids, and Suspensions 58 Summary TABLE 8.12 59 60 15 Learning Check Solution 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 61 Osmosis 62 Osmosis In osmosis, water (solvent) flows from the lower solute concentration into the 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? higher solute concentration. The level of the solution with the higher concentration rises. The concentrations of the two 4% starch 10% starch H 2O solutions become equal with time. semipermeable membrane 63 64 16 Water flow Equalizes Osmotic Pressure § The 10% starch solution is diluted by the flow of water out of the 4% and its volume increases. Osmotic pressure is § The 4% solution loses water and its volume decreases. § Produced by the solute particles dissolved in a § Eventually, the water flow between the two becomes equal. § Equal to the pressure that would prevent the flow of solution. additional water into the more concentrated solution. § Greater as the number of dissolved particles in the solution increases. 7% starch 7% starch H 2O 65 Learning Check 66 Solution A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below. 1. Solution ____ has the greater osmotic pressure. A semipermeable membrane separates a 10% starch solution (A) from a 5% starch solution (B). If starch is a colloid, fill in the blanks in the statements below. 1. Solution A has the greater osmotic pressure. 2. Water initially flows from ___ into ___. 2. Water initially flows from B into A. 3. The level of solution ____will be lower. 3. The level of solution B will be lower. 67 68 17 Osmotic Pressure of the Blood Isotonic Solutions Red blood cells 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. § 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. 69 Hypotonic Solutions 70 Hypertonic Solutions 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. § 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. § Causes crenation: RBCs shrinks in size. 71 72 18 Dialysis Learning Check In dialysis, Indicate if each of the following solutions is 1) isotonic 2) hypotonic 3) hypertonic § Solvent and small solute particles pass through an artificial membrane. A.____ 2% NaCl solution B.____ 1% glucose solution C.____ 0.5% NaCl solution D.____ 5% glucose solution § Large particles are retained inside. § Waste particles such as urea from blood are removed using hemodialysis (artificial kidney). 73 74 Solution Learning Check Indicate if each of the following solutions is When placed in each of the following, indicate if a red 1) isotonic 2) hypotonic blood cell will 3) hypertonic 1) not change 2) hemolyze A._3_ 2% NaCl solution A.____ 5% glucose solution B._2_ 1% glucose solution B.____ 1% glucose solution C._2_ 0.5% NaCl solution C.____ 0.5% NaCl solution D._1_ 5% glucose solution D.____ 2% NaCl solution 75 3) crenate 76 19 Solution Hemodialysis When placed in each of the following, indicate if a red § When kidneys fail, an artificial kidney uses hemodialysis to remove waste particles such as urea from blood. blood cell will 1) not change 2) hemolyze 3) crenate A._1_ 5% glucose solution B._2_ 1% glucose solution C._2_ 0.5% NaCl solution D._3_ 2% NaCl solution 77 78 Learning Check Solution Each of the following mixtures is placed in a dialyzing Each of the following mixtures is placed in a dialyzing bag and immersed in pure water. Which substance, if bag and immersed in pure water. Which substance, if any, will be found in the water outside the bag? any, will be found in the water outside the bag? A. 10% KCl solution A. 10% KCl solution B. 5% starch solution B. 5% starch solution None; starch is retained. KCl ( K+, Cl−) C. 5% NaCl and 5% starch solutions C. 5% NaCl and 5% starch solutions NaCl (Na+, Cl−), but starch is retained. 79 80 20