Aqueous Solutions Concentration / Calculations Dr. Ron 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.

Concentration and Temperature Relative Solution Concentrations: Saturated Unsaturated Supersaturated _____________ Dilute Concentrated A solution of 35g of potassium chloride in 100g H 2 25 o C is Saturated & 75 o C it is Unsaturated but Concentrated. Refer to: Lab Manual & Workshop

QUESTION What describes a solution of 0.250g SO 2 in 1.00L of H 2 10 o C? A) Dilute B) Concentrated C) Saturated D) Unsaturated

QUESTION What describes a solution of 0.250g SO 2 in 1.00L of H 2 70 o C? A) Dilute B) Concentrated C) Saturated D) Unsaturated

What describes a solution of 25.0g ammonium chloride in 0.10kg of H 2 10 o C? A) Dilute B) Concentrated C) Saturated D) Unsaturated

What describes a solution of 25.0g ammonium chloride in 0.10kg of H 2 70 o C? A) Dilute B) Concentrated C) Saturated D) Unsaturated

DHMO, dihydromonoxide : “The Universal” Solvent

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.

QUESTION

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”.

Salt dissolving in a glass of water Refer to Lab Manual & Workshop:

Water dissolving an ionic solid

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 %

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

QUESTION

HCl 1.0M 100% Ionized 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

QUESTION Solutions: molarity & volume  mass

Preparation of Solutions

Preparing a Standard Solution

QUESTION

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

QUESTION

Solution Dilution

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

Solution Applications Solution Applications Scoville Units / Capsaicin Capsaicin: C 18 H 27 NO 3 Will all molecules with the same molecular formula taste hot?

Molecular Applications Molecular Applications Capsaicin / Prosidol Capsaicin: C 18 H 27 NO 3 Will all molecules with the same molecular formula taste hot? Prosidol: C 18 H 27 NO 3 (opiate analgesic)

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

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

QUESTION

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

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

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

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

QUESTION