Solutions Chemistry 100 Solutions Solute and Solvent Solvent

Solutions Chemistry 100

Solutions: Solute and Solvent • Solvent: present in a much larger amount • Solute: present in a smaller amount • Solutions are homogeneous mixtures of two or more substances that forms when there is sufficient attraction between the particles of the solute and solvent • Solute and the solvent have similar polarities “like dissolve like”

Water as a Solvent • One of the most common solvents in nature • Polar molecule due to polar O–H bonds • Molecules form hydrogen bonds important in many biological compounds

Formation of Solutions “Like Dissolve Like” Polarity of solute and solvent must be similar to form a solution. There must be an attraction between the particles of the solute and solvent for there to be sufficient energy to form a solution.

Solutions with Ionic and Polar Solutes Ionic solutes: Na. Cl ions are hydrated in solution with many H 2 O molecules surrounding each ion. Strong solute-solvent attractions between Na+ and Cl- ions and polar H 2 O provide energy needed to form the solution.

Solutions with Nonpolar Solvents Nonpolar solvents • Are covalent compounds that contain many nonpolar bonds • Are often hydrocarbons, or molecules that contain almost no oxygen atoms • They do not dissolve in water or polar solvents

Electrolytes and Nonelectrolytes In water, • Strong Electrolytes produce ions and conduct an electric current • Weak Electrolytes produce a few ions • Nonelectrolytes do not produce ions

Strong Electrolytes • Dissociate completely in water, producing positive and negative ions • Conduct an electric current in water • In equations show the formation of ions in aqueous solutions

Weak Electrolyte • Dissociates only slightly in water • In water forms a solution of a few ions and mostly remain as molecules

Nonelectrolytes • Dissolve as molecules in water • Do Not produce ions in water • Do Not conduct an electric current

Equivalents • An equivalent (Eq) is the amount of that ion equal to 1 mole of positive or negative electrical charge (+ or ). • In any solution, the charge of the positive ions is always balanced by the charge of the negative ions.

Study Check 1 Calculate Equivalents 1. In 1 mole of Fe 3+, there are: A. 1 Eq B. 2 Eq C. 3 Eq 2. In 2. 5 moles of SO 42 , there are: A. 2. 5 Eq B. 5. 0 Eq C. 1. 0 Eq 3. An IV bottle contains Na. Cl. If the Na+ is 34 m. Eq/L, the Cl is: A. 34 m. Eq/L B. 0 m. Eq/L C. 68 m. Eq/L

Study Check 1 Answers 1. In 1 mole of Fe 3+, there are 3 Eq. 2. In 2. 5 moles of SO 42 , there are 2. 5 mole SO 42 2 Eq = 5. 0 Eq 1 mole SO 42 3. An IV bottle contains Na. Cl. If the Na+ is 34 m. Eq/L, the Cl is 34 m. Eq/L.

Solubility Amount of a solute that can dissolve in a given amount of solvent 3 Factors: Type of solute Type of solvent Temperature The solubility of most solids is greater as temperature increases. Expressed as grams of solute in 100 grams (or m. L) of solvent, usually water Solubility = g of solute 100 g water

Saturated Versus Unsaturated Solutions Unsaturated solutions • A solute readily dissolves when added • Contain less than the maximum amount of solute • More solute can be dissolved Saturated Solution • Contain the maximum amount of solute that can dissolve • Often have undissolved solute at the bottom of the container • Contain solute that dissolves as well as solute that Recrystallizes in an equilibrium process

Henry’s Law states that the solubility of a gas in a liquid is directly related to the pressure of that gas above the liquid. • At higher pressures, more gas molecules enter and dissolve in the liquid.

Solution Concentrations The concentration of a solution is the amount of solute dissolved in a specific amount of solution Concentration = amount of solute amount of solution • Solutions are commonly prepared based on a percent concentration • Mass Percent (% m/m) • Volume Percent (% v/v) • Mass/Volume Percent (% m/v) • Molarity (moles solute/liters solution)

Mass Percent (m/m) Concentration by mass of solute in a solution mass percent = g of solute 100 g of solute + g of solvent Amount in g of solute in 100 g of solution (conversion factor for mass percent) mass percent = g of solute 100 g of solution

Calculating Mass Percent A solution is prepared by mixing 15. 0 g of Na 2 CO 3 and 235 g of H 2 O. Calculate the mass percent (% m/m) of the solution. Determine the quantities of solute and solution. mass solute = 15. 0 g Na 2 CO 3 mass solution = 15. 0 g Na 2 CO 3 + 235 g H 2 O = 250. g Write the concentration expression. mass % (m/m) = g solute 100 g solution Substitute solute and solution quantities into the expression. mass % (m/m) = 15. 0 g Na 2 CO 3 100 = 6. 00% Na 2 CO 3 250. g solution

Volume Percent (v/v) Concentration Volume % (v/v) = m. L of solute m. L of solution 100 solute (m. L) in 100 m. L of solution (conversion factor for volume percent) volume % (v/v) = 35% = m. L of solute 100 m. L of solution 35 m. L solute 100 m. L of solution

Mass/Volume Percent (m/v) Concentration mass/volume % (m/v) = grams of solute m. L of solution x 100% Solute in 100 m. L of solution mass/volume % (m/v) = grams of solute 100 m. L of solution The mass/volume percent is widely used in hospitals for the preparation of intravenous solutions and medicines. 5% (m/v) glucose solution = 5 g glucose/ 100 m. L solution

Molarity (M) Concentration Molarity is a concentration term for solutions Molarity (M) = moles of solute liters of solution Preparing a 1. 0 M solution of Na. Cl: 58. 5 g (1 mole) to 1. 00 L of water M = 1. 0 mole Na. Cl = 1. 0 M Na. Cl solution 1 L Na. Cl

Molarity Calculations What is the molarity of 0. 500 L of Na. OH solution if it contains 6. 00 g of Na. OH? Determine the quantities of solute and solution. Solute: 1 mole of Na. OH = 40. 0 grams of Na. OH 6. 00 g Na. OH 1 mole Na. OH = 0. 150 mole of Na. OH 40. 0 g Na. OH The volume of the solution is 0. 500 L. M = moles solute = 0. 150 mole Na. OH = 0. 300 M Na. OH L solution 0. 500 L solution

Study Check 2 What is the molarity of 0. 225 L of a KNO 3 solution containing 34. 8 g of KNO 3?

Study Check 2 Answer What is the molarity of 0. 225 L of a KNO 3 solution containing 34. 8 g of KNO 3? M= moles solute L solution Determine the moles of solute and volume of solution. Solute: 1 mole of KNO 3 = 101. 1 grams of KNO 3 34. 8 g KNO 3 1 mole KNO 3 = 0. 344 mole of KNO 3 101. 1 g KNO 3 Solution: 0. 225 L M = moles solute = 0. 344 mole of KNO 3 = 1. 53 M KNO 3 L solution 0. 225 L solution

Dilution of Solutions • water is added • volume increases • mass of solute remains the same • concentration decreases C 1 V 1 = C 2 V 2 Molarity: M 1 V 1 = M 2 V 2

Dilution: Molarity What is the final concentration when 0. 50 L of 6. 0 M HCl solution is diluted to a final volume of 1. 0 L? Prepare a table of the concentrations and volumes of the solutions. C 1 = 6. 0 M V 1 = 0. 50 L C 2 = ? M V 2 = 1. 0 L Rearrange the dilution expression to solve for the unknown quantity. C 1 V 1 = V 2 C 2 M 1 x. V 1 = M 2 V 2 Substitute the known quantities into the dilution expression and solve. 6. 0 M 0. 50 L = 3. 0 M HCl 1. 0 L

Study Check 3 What is the molarity (M) of a solution prepared by diluting 0. 180 L of 0. 600 M HNO 3 to 0. 540 L?

Study Check 3 Answer Prepare a table of the concentrations and volumes of the solutions. M 1 = 0. 600 M V 1 = 0. 180 L M 2 = ? V 2 = 0. 540 L Rearrange the dilution expression to solve for the unknown quantity. M 1 V 1 = M 2 V 2 M 2 = M 1 V 1 V 2 Substitute the known quantities into the dilution expression and solve. M 2 = 0. 600 M x 0. 180 L = 0. 200 M solution 0. 540 L

Solutions, Colloids, Suspensions

Osmosis Water (solvent) move through a semipermeable membrane from the lower solute concentration into the higher solute concentration • the level of the solution with the higher solute concentration rises • the concentrations of the two solutions become equal with time Osmotic Pressure is equal to the pressure that would prevent the flow of additional water into the more concentrated solution • depends on the concentration of solute particles in the solution • greater as the number of dissolved particles in the solution increases produced by the solute particles dissolved in a solution

Isotonic, Hypertonic Solution

Dialysis Solvent and small solute particles pass through an artificial membrane • large particles are retained inside • waste products such as urea and excess water are removed from blood using hemodialysis (artificial kidney)