Slide 1Slide 2Slide 3Concentrations of SolutionsConcentrations of SolutionsClicker questionSlide 7Concentrations of SolutionsSlide 9Dilution of SolutionsDilution of SolutionsSlide 12Slide 13Using Solutions in Chemical ReactionsSlide 15Slide 16Slide 17Using Solutions in Chemical ReactionsSlide 19Slide 20Slide 211Calculate the mass of 8.00% w/w NaOH solution that contains 32.0 g of NaOH.Mass of solute x 100% = Mass of solution32.0 g / x g = .08x= 400 g solution2Calculate the mass of NaOH in 300.0 mL of an 8.00% w/w NaOH solution. Density is 1.09 g/mL.300D=m/v1.09 g/mL = m/300.0 Lm =327 g NaOhMass of solute / mass of solution x 100 = x g / 327 g = .08x = 26.2 g3What volume of 12.0% w/w KOH contains 40.0 g of KOH? The density of the solution is 1.11 g/mL.1.11 g/mL = 40.0 g / vv = 36.0 mL in 12.0%36.0 mL / 12.0% = x/ 100%x = 300. mL4Concentrations of SolutionsSecond common unit of concentration:mLmmolLmolessolution of liters ofnumber solute of moles ofnumber molarity MM5Concentrations of SolutionsCalculate the molarity of a solution that contains 12.5 g of sulfuric acid in 1.75 L of solution.Molarity = moles / liter12.5 g H2SO4 / 96.065 g/mol = .127 moles H2SO4M = .127 / 1.75 = .0728 M H2SO4Clicker questionWhat is the molarity of a solution that contain 10.0 moles of a solute in 2.00 l of solution?M = 10.0 moles / 2.00 L = 5.007Determine the mass of calcium nitrate required to prepare 3.50 L of a 0.800 M calcium nitrate..800 M = x moles / 3.50 L x = 2.80 moles Ca(NO3)2 x (164.1 g/mol)x = 459 g Ca(NO3)28Concentrations of Solutions•One of the reasons that molarity is commonly used is because:M x L = moles solute and M x mL = mmol solute9The specific gravity of a concentrated HCl is 1.185 and it is 36.31% w/w HCl. What is its molarity?Specific gravite = density of substance/density of waterDensity = 1.185 g/ml HCl* 36.31 g / 100.0g = .43027 g/mol.43027 g/mol (1000 mol/L)= 430.27 g/L / 36.46 g/mol = 11.8 moles/L HCl10Dilution of Solutions•To dilute a solution, add solvent to a concentrated solution.–One method to make tea “less sweet.”–How fountain drinks are made from syrup.•The number of moles of solute in the two solutions remains constant.•The relationship M1V1 = M2V2 is appropriate for dilutions, but not for chemical reactions.11Dilution of Solutions•Common method to dilute a solution involves the use of volumetric flask, pipet, and suction bulb.12If 10.0 mL of 12.0 M HCl is added to enough water to give 100. mL of solution, what is the concentration of the solution?M1V1 = M2V212.0 M (.0100 L) = M2 (.100L)M2 = 1.2 M13Clicker questionWhat volume of an 18.0 M sulfuric acid is required to make 2.50 L of a 2.40 M sulfuric acid solution?18.0 M ( V1) = 2.40 M(2.50 L)V1 = .333 L14Using Solutions in Chemical ReactionsCombine the concepts of molarity and stoichiometry to determine the amounts of reactants and products involved in reactions in solution.15What volume of a 0.500 M BaCl2 is required to completely react with 4.32 g of Na2SO4?4.32 g Na2SO4 / (142.05 g/mol) = .0304 mol.0304 mol Na2SO4 (1 mol BaCl2 / 1 mol Na2SO4)= .0304 mol BaCL2.500 M = .0304 mol / VV = .0608 L BaCl2 = 60.8 mol .500 M BaCl2 NaCl 2 + BaSO BaCl + SONa424216(a) What volume of a 0.200 M NaOH will react with 50.0 mL of a 0.200 M aluminum nitrate, Al(NO3)3?50.0 mL (1L / 1000 mL) = 0.050 L.200 M = x moles / .050 L x = 0.010 moles Al(NO3)3.01 mol Al(NO3)3 (3 mol NaOH / 1 mol Al(NO3)3)=0 .030 moles NaOH.0300 moles NaOH / .200 moles / L=0 .150 L = 150 mL of .200 M NaOH( ) ( )3 33 3Al NO 3 NaOH Al OH 3 NaNO+ � +17(b) What mass of Al(OH)3 precipitates in (a)?0.0100 mol Al(NO3)3 ( 1 mol Al(OH)3 / 1 mol Al(NO3)3)= 0.0100 mol Al(OH)3.0100 mol Al(OH)3 / 77.98 g/mol = 0.780 g Al(OH)318Using Solutions in Chemical Reactions•Titrations are a method of determining the concentration of an unknown solutions from the known concentration of a solution and solution reaction stoichiometry.–Requires special lab glassware•Buret, pipet, and flasks –Must have an indicator also19What is the molarity of a KOH solution if 38.7 mL of the KOH solution is required to react with 43.2 mL of a 0.223 M HCl?2KOH + HCl KCl + H O�Clicker question# of moles? 44.1 mL of a 0.103 M HCl21What is the molarity of a barium hydroxide solution if 44.1 mL of a 0.103 M HCl is required to react with 38.3 mL of the Ba(OH)2 solution?2 2 2Ba(OH) + 2 HCl BaCl + 2 H