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© Copyright 1995-2010 R.J. Rusay Aqueous Solutions Concentration / Calculations Dr. Ron Rusay Spring 2008.

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Presentation on theme: "© Copyright 1995-2010 R.J. Rusay Aqueous Solutions Concentration / Calculations Dr. Ron Rusay Spring 2008."— Presentation transcript:

1 © Copyright R.J. Rusay Aqueous Solutions Concentration / Calculations Dr. Ron Rusay Spring 2008

2 © Copyright R.J. Rusay Solutions  Homogeneous solutions are comprised of solute(s), the substance(s) dissolved, [The lesser amount of the component(s) in the mixture], and  solvent, the substance present in the largest amount.  Solutions with less solute dissolved than is physically possible are referred to as “unsaturated”. Those with a maximum amount of solute are “saturated”.  Occasionally there are extraordinary solutions that are “supersaturated” with more solute than normal.

3 © Copyright R.J. Rusay Concentration and Temperature Relative Solution Concentrations: Saturated Unsaturated Supersaturated Reading: Lab Manual pp & Workshop: pp

4 © Copyright R.J. Rusay DHMO, dihydromonoxide : “The Universal” Solvent

5 © Copyright R.J. Rusay Water : “The Universal” Solvent Generally, likes dissolve likes, i.e. polar-polar and nonpolar-nonpolar. If polar and nonpolar mix, eg. oil and water: The oil (nonpolar) and water (polar) mixture don’t mix and are immiscible. If liquids form a homogeneous mixture, they are miscible.

6 QUESTION

7 © Copyright R.J. Rusay Aqueous Reactions & Solutions  Many reactions are done in a homogeneous liquid or gas phase which generally improves reaction rates.  The prime medium for many inorganic reactions is water which serves as a solvent (the substance present in the larger amount), but does not react itself.  The substance(s) dissolved in the solvent is (are) the solute(s). Together they comprise a solution. The reactants would be the solutes.  Reaction solutions typically have less solute dissolved than is possible and are “unsaturated”.

8 © Copyright R.J. Rusay Salt dissolving in a glass of water Refer to Lab Manual & Workshop:

9 © Copyright R.J. Rusay Water dissolving an ionic solid

10 © Copyright R.J. Rusay Solution Concentrations Refer to Lab Manual  A solution’s concentration is the measure of the amount of solute dissolved.  Concentration is expressed in several ways. One way is mass percent. Mass % = Mass solute / [Mass solute + Mass solvent ] x100  What is the mass % of 65.0 g of glucose dissolved in 135 g of water? Mass % = 65.0 g / [ ]g x 100 = 32.5 % = 32.5 %

11 © Copyright R.J. Rusay Solution Concentration  Concentration is expressed more importantly as molarity (M). Molarity (M) = Moles solute / Liter Solution  An important relationship is M x V solution = mol  This relationship can be used directly in mass calculations of chemical reactions.  What is the molarity of a solution of 1.00 g KCl in 75.0 mL of solution? M KCl = [1.00g KCl / 75.0mL ][1mol KCl / g KCl ][ 1000mL / L] = 0.18 mol KCl / L

12 QUESTION

13 © Copyright R.J. Rusay Figure 4.6 HCl 1.0M 100% Ionized Figure 4.8 Acetic Acid (HC 2 H 3 O 2 ) < 100% Ionized Refer to Lab Manual Solution Concentrations: Solute vs. Ion Concentrations [H+] = [Cl-] = 1.0M[H+] = [C 2 H 3 O 2 -] < 1.0M

14 QUESTION Solutions: molarity & volume  mass

15 © Copyright R.J. Rusay Preparation of Solutions

16 © Copyright R.J. Rusay Preparing a Standard Solution

17 QUESTION

18 © Copyright R.J. Rusay Solution Concentration  The following formula can be used in dilution calculations: M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 V 2  A concentrated stock solution is much easier to prepare and then dilute rather than preparing a dilute solution directly. Concentrated sulfuric acid is 18.0M. What volume would be needed to prepare 250.mL of a 1.50M solution?  V 1 = M 2 V 2 / M 1  V 1 = 1.50 M x 250. mL / 18.0 M  V 1 = 20.8 mL

19 QUESTION

20 © Copyright R.J. Rusay Solution Dilution

21 © Copyright R.J. Rusay Solution Applications A solution of barium chloride was prepared by dissolving g in water to make mL of solution. What is the concentration of the barium chloride solution? M BaCl2 = ? M BaCl2 M BaCl2 = = [ g BaCl2 / mL ][1mol BaCl2 / g BaCl2 ] [1000mL / L] = mol / L

22 What happens to the number of moles of C 12 H 22 O 11 (sucrose) when a 0.20 M solution is diluted to a final concentration of 0.10 M? A) The number of moles of C 12 H 22 O 11 decreases. B) The number of moles of C 12 H 22 O 11 increases. C) The number of moles of C 12 H 22 O 11 does not change. D) There is insufficient information to answer the question. QUESTION

23 © Copyright R.J. Rusay Solution Applications mL of this solution was diluted to make exactly mL of solution which was then used to react with a solution of potassium sulfate. What is the concentration of the diluted solution. M 2 = ? M BaCl2 M 1 M BaCl2 = M 1 M 2 = M 1 V 1 / V 2 M 2 = M x mL / mL M 2 = M

24 QUESTION

25 © Copyright R.J. Rusay Solution Applications mL of a barium chloride solution required mL of the potassium sulfate solution to react completely. M K2SO4 = ? mL of a M 2 = M barium chloride solution required mL of the potassium sulfate solution to react completely. M K2SO4 = ? BaCl 2 (aq) + K 2 SO 4 (aq)  ? + ? BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) ?M K2SO4 K2SO4 ] [? mol K2SO4 / ? mol BaCl2 ] ?M K2SO4 = [M BaCl2 x V BaCl2 / V K2SO4 ] [? mol K2SO4 / ? mol BaCl2 ] 1 mol K2SO mol BaCl2 x L BaCl2 x 1 mol K2SO4 L BaCl L K2SO4 1 mol BaCl2 L BaCl2 x L K2SO4 x 1 mol BaCl2 ?M K2SO4 ?M K2SO4 = ?M K2SO K2SO4 / L K2SO K2SO4 ?M K2SO4 = mol K2SO4 / L K2SO4 = M K2SO4

26 © Copyright R.J. Rusay Solution Applications How many grams of potassium chloride are produced? BaCl 2 (aq) + K 2 SO 4 (aq)  ? + ? BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) ?g KCl = mol BaCl2 / L BaCl2 x L BaCl2 X 2 mol KCl / 1 mol BaCl2 X g KCl /mol KCl g KCl = g KCl

27 © Copyright R.J. Rusay Solution Applications If mL of a 0.10 M solution of barium chloride was reacted with mL of a 0.20 M solution of potassium sulfate, what would be the theoretical yield of barium sulfate? BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) Mol BaCl2 = M BaCl2 x V BaCl2 = 0.10 mol BaCl2 / L BaCl2 x L BaCl2 1 mol BaCl2 = 2.0 x Mol K2SO4 = M K2SO4 x V K2SO4 = 0.20 mol K2SO4 / L K2SO4 x L K2SO4 1 mol K2SO4 = 3.0 x Which is the Limiting Reagent? 2.0 x < 3.0 x x mol is limiting

28 © Copyright R.J. Rusay Solution Applications If mL of a 0.10 M solution of barium chloride was reacted with mL of a 0.20 M solution of potassium sulfate, what would be the theoretical yield of barium sulfate? BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) BaCl 2 (aq) + K 2 SO 4 (aq)  2 KCl (aq) + BaSO 4 (s) Must use the limiting reagent: 0.10 mol BaCl2 x L BaCl2 x 1 mol BaSO4 x g BaSO4 L BaCl2 1 mol BaCl2 mol BaSO4 = = 0.47 g

29 QUESTION


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