Solutions Dr. Ron Rusay Spring 2003

Limestone Caves: Solubility of CaCO 3

Solutions  Substances can mix together to form homogeneous mixtures (solutions). The material present in the larger amount is the solvent and the other(s) is (are) the solute(s). Together they form a solution.  The most common solutions are liquids. The solute can be a solid, liquid or gas which is dissolved in a liquid solvent.  The most common solvent is water. © Copyright R.J. Rusay

DHMO, dihydromonoxide : “The Universal” Solvent

Water as a 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.

Salt dissolving in a glass of water

Water dissolving an ionic solid

Concentration and Temperature Besides amount, the rate also increases. What are two other ways of increasing the solubility of a solid, eg. sugar in coffee? Relative Solution Concentrations: Saturated Unsaturated Supersaturated

Solution Types  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.

A Giant, Single Crystal and Nuclear Energy

Tooth Enamel (Dentyne) & Fluoride Treatment

Gas Solubility P 2 = P 1 x [P 2 / P 1 ] P is the partial pressure of the gas vapor. Solubility units (Concentration) are usually: g / 100 ml

Preparation of Solutions

Solution Concentration  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 x100 = 32.5 % = 32.5 % © 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 = 1.00g KCl x 1mol KCl / g KCl x 1/ 75mL x 1000mL / L = 0.18 mol / L © Copyright R.J. Rusay

Acid-Base Titration

Solution Applications Neutralization-Titration mL of vinegar, a solution of acetic acid (aa), required mL of a M (mol/L) solution of a sodium hydroxide solution to react completely. M aa = ? HC 2 H 3 O 2 (aq) + NaOH (aq)  ? + ? HC 2 H 3 O 2 (aq) + NaOH(aq)  1 NaC 2 H 3 O 2 (aq) + 1H 2 O (l) ?M aa HC2H3O2 ] [? mol HC2H3O2 / ? mol NaOH ] ?M aa = [M NaOH x V NaOH / V HC2H3O2 ] [? mol HC2H3O2 / ? mol NaOH ] NaOH 1 mol HC2H3O mol NaOH x L NaOH x 1 mol HC2H3O2 L NaOH L HC2H3O2 1 mol NaOH L NaOH x L HC2H3O2 x 1 mol NaOH ?M aa ?M aa = ?M aa HC2H3O2 / L HC2H3O HC2H3O2 ?M aa = mol HC2H3O2 / L HC2H3O2 = M HC2H3O2

Solution Applications What is the weight percent of the acetic acid (aa)? ( ( Density of vinegar = g/mL; Molar Mass (aa) = MW (aa) in grams/ mole aa = g aa/mole aa) � (Molarity aa x Volume aa) x (60.056g aa /mole aa) � = mass aa = g aa [(mass of Acetic Acid) / (mass of vinegar) ] * 100% = % Acetic Acid � mass of vinegar = density x sample volume = g � % = g aa / g * 100 = 2.3%

Solution Dilution