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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
Molarity
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 litersMolarity (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/molSo: 10.0g x (1 mol/35.45g)=0.282 mol NaClMolar concentration: 0.282 mol/1.00 L = 0.282 M
Molarity – Moles - Volume
• 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
moles of solute
volume of solution in litersMolarity (M) =
mol
Volume (L)Molarity (M) =
Practice
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
mol HCl =0.10 M HCl x 2.5 L 0.10 mol HCl=1 L
x 2.5 L
= 0.25 mol HCl
= 0.10 mol/1 L HCl
PracticeWhat 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
Vol soln = 0.50 mol NaOH0.10 M NaOH
= 0.50 mol NaOH0.10 mol NaOH
1L
= 0.50 mol NaOH X 1L0.10 mol NaOH = 5 L
More PracticeHow 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
– 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
0.15 mol
25 mL HCl
+
25 mL H2O
=50 mL0.15 mol
Dilutionsmoles 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
Molarity = moles solute
Volume of solution
Dilution Calculation
• 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
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 Vcx Mc= Vdx Md Solve for Md
Note: Vcand Vd do not have to be in liters, but they must be in the same units.
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 Md = 0.10MVd = 250 mL
Vc = 10.0 mLMd = ? 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.
Molarity
B
Molar
mass
Molarity
A
Solution Stoichiometry
gramsB
gramsA
molesA
molesB
VolSoln A
Vol Soln B
Molar
mass
Molar ratio
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)
Molarity
NaOH
Approach
molesNaOH
molesH3PO4
VolNaOH Soln
Molar ratio
25.0 mL NaOH soln
0.025 L NaOH soln
2.5 M (=mol/L)
3 mol NaOH/1 mol H3PO4
Mol NaOH = 25.0 mL x 1000 mL
1L
1 L
2.5 molx = 0.0625 mol NaOH
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)
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.
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.