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.

Concentrations of Solutions Concentration of a solution: the more solute in a given volume of solvent, the more concentrated 1 tsp salt (NaCl)/cup of water vs 3 Tbsp salt/cup water

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 = M moles of solute volume of solution in liters Molarity (M) =

Preparing a 1.0 Molar Solution One liter of a 1.00 M NaCl solution need 1.00 mol of NaCl weigh out 58.5 g NaCl (1.00 mole) and add water to make 1.00 liter (total volume) of solution. Copyright © 2005 by Pearson Education, Inc. Publishing as Benjamin Cummings

Molarity Practice What 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 M 0.5 mol NaCl/0.50 L = 0.5/0.50 mol/L = 1 mol/L = 1 M

Molarity Practice What is the molar NaCl concentration if you have 10.0 g of NaCl in 1.00 L of solution? Have grams not mols! Grams  mol Need molar mass NaCl: 23.00 + 35.45= 58.45 g/mol So: 10.0g x (1 mol/35.45g)=0.282 mol NaCl Molar concentration: 0.282 mol/1.00 L = 0.282 M

Molarity – Moles - Volume moles of solute volume of solution in liters Molarity (M) = mol Volume (L) Molarity (M) = Have mol and vol  molarity Have molarity and vol  mol of solute Have molarity and mol of solute  volume AND: mol of solute  grams of solute

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 soln 0.10M HCl Find: mol HCl Use: mol = molarity x volume = 0.10 mol/1 L HCl 0.10 mol HCl mol HCl = 0.10 M HCl x 2.5 L = x 2.5 L 1 L = 0.25 mol HCl

Practice What volume of a 0.10 M NaOH solution is needed to provide 0.50 mol of NaOH? Given: 0.50 mol NaOH 0.10 M NaOH Find: vol soln Use: vol soln = mol solute / molarity = 0.10 mol NaOH / 1L 0.50 mol NaOH 0.50 mol NaOH Vol soln = = 0.10 M NaOH 0.10 mol NaOH 1L 0.50 mol NaOH X 1L = = 5 L 0.10 mol NaOH

More Practice How many grams of CuSO4 are needed to prepare 250.0 mL of 1.00 M CuSO4? Given: 250.0 mL soln 1.00 M CuSO4 Find: g CuSO4 Use: mol CuSO4 = molarity x volume Molarity = mol / 1L Vol = 250.0 mL

Concentration of Solutions Interconverting Molarity, Moles, and Volume g CuSO4 = 250.0 mL soln x 1 L x 1.00 mol 1000 mL 1 L soln x 159.6 g CuSO4 1 mol = 39.9 g CuSO4

Steps involved in preparing solutions from pure solids

Steps involved in preparing solutions from pure solids Calculate the amount of solid required Weigh out the solid Place in an appropriate volumetric flask Fill 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.

Dilutions Many laboratory chemicals such as acids are purchased as concentrated solutions (stock solutions). e.g. 12 M HCl 12 M H2SO4 More dilute solutions are prepared by taking a certain quantity of the stock solution and diluting it with water.

Dilutions A 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 HCl 25 mL HCl 25 mL H2O 0.15 mol 50 mL 0.15 mol + =

moles solute = moles solute Dilutions moles solute = moles solute before dilution after dilution Although the number of moles of solute does not change, the volume of solution does change. The concentration of the solution will change since moles solute Molarity = Volume of solution

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 Vd where Mc= initial concentration (mol/L)= more concentrated Vc = initial volume of more conc. solution Md = final concentration (mol/L) in dilution Vd = final volume of diluted solution

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 mL Mc = 6.00 M Vd = 50.0 mL Find: Md Use Note: Vcand Vd do not have to be in liters, but they must be in the same units. Vcx Mc= Vdx Md Solve for Md

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.

Practice How 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 = 5.0M Vd = 250 mL Md = 0.10M Md = ? Vc = 10.0 mL Mc = 10.0M Vd = 250 mL

Solution Stoichiometry Remember: reactions occur on a mole to mole basis. For pure reactants, we measure reactants using mass For reactants that are added to a reaction as aqueous solutions, we measure the reactants using volume of solution.

Solution Stoichiometry grams A Molar mass moles A Molarity A Vol Soln A Molar ratio Molar mass grams B moles B Molarity B Vol Soln B

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: 25.0 mL 2.5 M NaOH balanced eqn: 3 mol NaOH/1 mol H3PO4 Find: moles of H3PO4 3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

Approach moles NaOH Vol NaOH Soln moles H3PO4 Molarity NaOH 2.5 M (=mol/L) Molar ratio 0.025 L NaOH soln 3 mol NaOH/1 mol H3PO4 moles H3PO4 25.0 mL NaOH soln 1L 2.5 mol Mol NaOH = 25.0 mL x x = 0.0625 mol NaOH 1000 mL 1 L

More practice What mass of aluminum hydroxide is needed to neutralize 12.5 mL of 0.50 M sulfuric acid?

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)

Titration

Practice If 35.50 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: 35.50 mL 2.5 M NaOH 50.0 mL of H3PO4 sol’n Find: molarity (mol/L) H3PO4 3NaOH (aq) + H3PO4 (aq)  Na3PO4 (aq) + 3H2O(l)

Strategy: M = moles L To 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.

Plan Molarity NaOH moles NaOH Vol NaOH Soln Molar ratio moles H3PO4

Mol H3PO4 = 35.5 mL x 1 L 1000 mL x 2.50 mol NaOH x 1 mol H3PO4 1 L 3 mol NaOH = 0.0296 mol H3PO4 We’re not done….we need molarity.

Molarity of H3PO4 Molarity = moles L = 0.0296 mol H3PO4 x 1000 mL 50.0 mL L = 0.592 M H3PO4