Presentation on theme: "Concentrations of Solutions"— Presentation transcript:
1 Concentrations of Solutions Behavior of solutions depend on compound itself and on how much is present, i.e. on the concentration.Two solutions can contain the same compounds but behave quite different because the proportions of those compounds are different.
2 Concentrations of Solutions Concentration of a solution: the more solute in a given volume of solvent, the more concentrated1 tsp salt (NaCl)/cup of watervs3 Tbsp salt/cup water
3 Units of molarity are: mol/L = M Molarity is one way to measure the concentration of a solution.A 1.00 molar (1.00 M) solution contains 1.00 mol solute in every 1 liter of solution.Units of molarity are: mol/L = Mmoles of solutevolume of solution in litersMolarity (M) =
5 Molarity PracticeWhat is the molar NaCl concentration if you have 0.5 mol NaCl in 1.00 L of solution?What is the molar NaCl concentration if you have 0.5 mol NaCl in 0.50 L of solution?0.5 mol NaCl/1.00 L = 0.5 mol/L = 0.5 M0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L = 1 mol/L = 1 M
6 Molarity PracticeWhat is the molar NaCl concentration if you have 10.0 g of NaCl in 1.00 L of solution?Have grams not mols!Grams molNeed molar massNaCl: = g/molSo: 10.0g x (1 mol/35.45g)=0.282 mol NaClMolar concentration: mol/1.00 L = M
7 Molarity – Moles - Volume moles of solutevolume of solution in litersMolarity (M) =molVolume (L)Molarity (M) =Have mol and vol molarityHave molarity and vol mol of soluteHave molarity and mol of solute volumeAND: mol of solute grams of solute
8 Practice 0.10 mol HCl mol HCl = 0.10 M HCl x 2.5 L = x 2.5 L 1 L How many moles of HCl are present in 2.5 L of 0.10 M HCl?Given: 2.5 L of soln0.10M HClFind: mol HClUse: mol = molarity x volume= 0.10 mol/1 L HCl0.10 mol HClmol HCl =0.10 M HCl x 2.5 L=x 2.5 L1 L= 0.25 mol HCl
9 PracticeWhat volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH?Given: mol NaOH0.10 M NaOHFind: vol solnUse: vol soln = mol solute / molarity= 0.10 mol NaOH / 1L0.50 mol NaOH0.50 mol NaOHVol soln ==0.10 M NaOH0.10 mol NaOH1L0.50 mol NaOHX 1L==5 L0.10 mol NaOH
10 More PracticeHow many grams of CuSO4 are needed to prepare mL of 1.00 M CuSO4?Given: mL soln1.00 M CuSO4Find: g CuSO4Use: mol CuSO4 = molarity x volumeMolarity = mol / 1LVol = mL
11 Concentration of Solutions Interconverting Molarity, Moles, and Volume g CuSO4 = mL soln x 1 L x 1.00 mol1000 mL 1 L solnx g CuSO41 mol= 39.9 g CuSO4
12 Steps involved in preparing solutions from pure solids
13 Steps involved in preparing solutions from pure solids Calculate the amount of solid requiredWeigh out the solidPlace in an appropriate volumetric flaskFill flask about half full with water and mix.Fill to the mark with water and invert to mix.You should be able to describe this process (including calculating the mass of solid to use) for any solution I specify.
14 DilutionsMany laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions).e.g. 12 M HCl12 M H2SO4More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.
15 DilutionsA given volume of a stock solution contains a specific number of moles of solute.e.g.: 25 mL of 6.0 M HCl contains 0.15 mol HCl(How do you know this???)If 25 mL of 6.0 M HCl is diluted with 25 mL of water, the number of moles of HCl present does not change.Still contains 0.15 mol HCl25 mL HCl25 mL H2O0.15 mol50 mL0.15 mol+=
16 moles solute = moles solute Dilutionsmoles solute = moles solutebefore dilution after dilutionAlthough the number of moles of solute does not change, the volume of solution does change.The concentration of the solution will change sincemoles soluteMolarity =Volume of solution
17 Dilution Calculation Mc x Vc = Md x Vd When a solution is diluted, the concentration of the new solution can be found using:Mc x Vc = Md x Vdwhere Mc= initial concentration (mol/L)= more concentratedVc = initial volume of more conc. solutionMd = final concentration (mol/L) in dilutionVd = final volume of diluted solution
18 Dilution Calculation Find: Md Use What is the concentration of a solution prepared by diluting 25.0 mL of 6.00 M HCl to a total volume of 50.0 mL?Given: Vc = 25.0 mLMc = 6.00 MVd = 50.0 mLFind: MdUseNote: Vcand Vd do not have to be in liters, but they must be in the same units.Vcx Mc= Vdx MdSolve for Md
19 Dilution Make a diluted solution once you know Vc and Vd Use a pipet to deliver a volume of the concentrated solution to a new volumetric flask.Add solvent to the line on the neck of the new flask.Mix well.
20 PracticeHow many mL of 5.0 M K2Cr2O7 solution must be diluted to prepare 250 mL of 0.10 M solution?If 10.0 mL of a 10.0 M stock solution of NaOH is diluted to 250 mL, what is the concentration of the resulting solution?Vc = ?Mc = 5.0MVd = 250 mLMd = 0.10MMd = ?Vc = 10.0 mLMc = 10.0MVd = 250 mL
21 Solution Stoichiometry Remember: reactions occur on a mole to mole basis.For pure reactants, we measure reactants using massFor reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.
22 Solution Stoichiometry gramsAMolarmassmolesAMolarityAVolSoln AMolar ratioMolarmassgramsBmolesBMolarityBVol Soln B
23 Solution Stoichiometry Practice If 25.0 mL of 2.5 M NaOH are needed to neutralize (i.e. react completely with) a solution of H3PO4, how many moles of H3PO4 were present in the solution?Given: mL 2.5 M NaOHbalanced eqn: 3 mol NaOH/1 mol H3PO4Find: moles of H3PO43NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3H2O(l)
24 Approach moles NaOH Vol NaOH Soln moles H3PO4 Molarity NaOH 2.5 M (=mol/L)Molar ratio0.025 L NaOH soln3 mol NaOH/1 mol H3PO4molesH3PO425.0 mL NaOH soln1L2.5 molMol NaOH = 25.0 mL xx= mol NaOH1000 mL1 L
25 More practiceWhat mass of aluminum hydroxide is needed to neutralize 12.5 mL of 0.50 M sulfuric acid?
26 Solution Stoichiometry Solution stoichiometry can be used to determine the concentration of aqueous solutions used in reactions.Concentration of an acid can be determined using a process called titration.reacting a known volume of the acid with a known volume of a standard base solution (i.e. a base whose concentration is known)
28 PracticeIf mL of 2.5 M NaOH are needed to neutralize 50.0 mL of an H3PO4 solution, what is the concentration (molarity) of the H3PO4 solution?Given: mL 2.5 M NaOH50.0 mL of H3PO4 sol’nFind: molarity (mol/L) H3PO43NaOH (aq) + H3PO4 (aq) Na3PO4 (aq) + 3H2O(l)
29 Strategy: M = molesLTo find the concentration of H3PO4 soln, we need both # moles and volume of H3PO4.Since volume is given, we can simply find moles and plug into the equation for M.