Chapter 8 Solutions 8.1 Solutions

Solutions: Solute and Solvent are homogeneous mixtures of two or more substances consist of a solvent and one or more solutes

Nature of Solutes in Solutions spread evenly throughout the solution cannot be separated by filtration can be separated by evaporation are not visible but can give a color to the solution

Examples of Solutions The solute and solvent in a solution can be a solid, liquid, and/or a gas.

Water Water is the most common solvent is a polar molecule forms hydrogen bonds between the hydrogen atom in one molecule and the oxygen atom in a different water molecule

Combinations of Solutes and Solvents in Solutions

Formation of a Solution Na+ and Cl– ions on the surface of a NaCl crystal are attracted to polar water molecules are hydrated in solution by many H2O molecules surrounding each ion

Equations for Solution Formation When NaCl(s) dissolves in water, the reaction can be written as H2O NaCl(s) Na+(aq) + Cl- (aq) solid separation of ions

Like Dissolves Like Two substances form a solution when there is an attraction between the particles of the solute and solvent when a polar solvent (such as water) dissolves polar solutes (such as sugar) and/or ionic solutes (such as NaCl) when a nonpolar solvent such as hexane (C6H14) dissolves nonpolar solutes such as oil or grease

Water and a Polar Solute

Like Dissolves Like Solvents Solutes Water (polar) Ni(NO3)2 CH2Cl2 (nonpolar) (polar) I2 (nonpolar)

8.2 Electrolytes and Nonelectrolytes Chapter 8 Solutions 8.2 Electrolytes and Nonelectrolytes

Solutes and Ionic Charge In water, strong electrolytes produce ions and conduct an electric current weak electrolytes produce a few ions nonelectrolytes do not produce ions

Strong Electrolytes Strong electrolytes dissociate in water, producing positive and negative ions dissolved in water will conduct an electric current in equations show the formation of ions in aqueous (aq) solutions H2O 100% ions NaCl(s) Na+(aq) + Cl(aq) H2O CaBr2(s) Ca2+(aq) + 2Br(aq)

Weak Electrolytes A weak electrolyte dissociates only slightly in water in water forms a solution of a few ions and mostly undissociated molecules HF(g) + H2O(l) H3O+(aq) + F(aq) NH3(g) + H2O(l) NH4+(aq) + OH(aq)

Nonelectrolytes Nonelectrolytes dissolve as molecules in water do not produce ions in water do not conduct an electric current

Classification of Solutes in Aqueous Solutions

Equivalents An equivalent (Eq) is the amount of an electrolyte or an ion that provides 1 mole of electrical charge (+ or –). 1 mole Na+ = 1 Eq 1 mole Cl− = 1 Eq 1 mole Ca2+ = 2 Eq 1 mole Fe3+ = 3 Eq

Electrolytes in IV Solutions

Electrolytes in Body Fluids In replacement solutions for body fluids, the electrolytes are given in milliequivalents/L (mEq/L). Ringer’s Solution Cations Anion_____ Na+ 147 mEq/L Cl− 155 mEq/L K+ 4 mEq/L Ca2+ 4 mEq/L 155 mEq/L = 155 mEq/L

Chapter 8 Solutions 8.3 Solubility

Solubility Solubility is the maximum amount of solute that dissolves in a specific amount of solvent expressed as grams of solute in 100 grams of solvent (usually water): g of solute 100 g water

Unsaturated Solutions contain less than the maximum amount of solute can dissolve more solute Dissolved solute

Saturated Solutions Saturated solutions contain the maximum amount of solute that can dissolve some undissolved solute at the bottom of the container Dissolved solute Undissolved solute

Effect of Temperature on Solubility depends on temperature of most solids increases as temperature increases of gases decreases as temperature increases

Solubility and Pressure Henry’s law states: 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 dissolve in the liquid

Soluble and Insoluble Salts Ionic compounds that dissolve in water are soluble salts do not dissolve in water are insoluble salts

Solubility Rules Soluble salts typically contain at least one ion from Groups 1A(1), NO3−, or C2H3O2− (acetate). Most other combinations are insoluble.

Examples of Using the Solubility Rules

Formation of a Solid When solutions of salts are mixed, a solid forms if ions of an insoluble salt are present Example: AgNO3(aq) + NaCl(aq) AgCl(s) + NaNO3(aq)

Equations for Forming Solids A full equation shows the formulas of the compounds. Pb(NO3)(aq) + 2NaCl(aq) PbCl2(s) + 2NaNO3(aq) An ionic equation shows the ions of the compounds. Pb2+(aq) + 2NO3−(aq) + 2Na+(aq) + 2Cl−(aq) PbCl2(s) + 2Na+(aq) + 2NO3−(aq) A net ionic equation shows only the ions that form a solid. Pb2+(aq) + 2Cl−(aq) PbCl2(s)

Guide to Writing New Ionic Equations for an Insoluble Salt

Finding the Insoluble Salt STEP 1 Write the ions of the reactants. Ba2+(aq) + 2NO3−(aq) + 2Na+(aq) + CO32−(aq) STEP 2 Write the new combinations of the ions and determine if an insoluble salt forms. BaCO3(s) + 2Na+(aq) + 2NO3−(aq) STEP 3 Write the ionic equation, including the insoluble salt as a solid in the products. Ba2+(aq) + 2NO3−(aq) + 2Na+(aq) + CO32 −(aq) BaCO3(s) + 2NO3−(aq) + 2Na+(aq) STEP 4 Write the net ionic equation deleting spectator ions. Ba2+(aq) + CO32−(aq) BaCO3(s)

8.4 Percent Concentration Chapter 8 Solutions 8.4 Percent Concentration

Concentration The concentration of a solution is the amount of solute dissolved in a specific amount of solution concentration = amount of solute amount of solution

Mass Percent Concentration Mass percent (m/m) concentration is the percent by mass of solute in a solution calculated using the formula: Mass = mass of solute(g) x 100% percent mass of solute(g) + mass of solvent(g) Mass percent = mass of solute(g) x 100% mass of solution(g)

Mass of Solution

Calculating Mass Percent The calculation of mass percent (m/m) requires the grams of solute (g of KCl) and grams of solution (g of KCl solution) g of KCl = 8.00 g g of solvent (water) = 42.00 g g of KCl solution = 50.00 g 8.00 g of KCl (solute) x 100% = 16.0% (m/m) 50.00 g of KCl solution

Guide to Calculating Solution Concentrations

Volume Percent The volume percent (v/v) is percent by volume of solute (liquid) in a solution volume % (v/v) = volume of solute x 100% volume of solution volume solute (mL) in 100 mL of solution. volume % (v/v) = mL of solute 100. mL of solution

Mass/Volume Percent The mass/volume percent (m/v) is mass/volume % (m/v) = grams (g) of solute x 100% volume(mL) of solution mass of solute(g) in 100 mL solution = g of solute 100. mL of solution

Preparation of a Solution: Mass/Volume Percent

Percent Conversion Factors Two conversion factors can be written for each type of percent concentration.

Guide to Using Concentration to Calculate Mass or Volume

Example of Using Percent Concentration (m/m) Factors How many grams of NaCl are needed to prepare 225 g of a 10.0% (m/m) NaCl solution? STEP 1 Given: 225 g of solution; 10.0% (m/m) NaCl Need: g of NaCl (solute) STEP 2 Plan: g of solution g of NaCl 45

Solution STEP 3 Write equalities and conversion factors: 10.0 g of NaCl = 100 g of NaCl solution 10.0 g NaCl and 100 g NaCl solution 100 g NaCl solution 10.0 g NaCl STEP 4 Set up problem to cancel grams of solution: 225 g NaCl solution x 10.0 g NaCl = 22.5 g of NaCl 100 g NaCl solution

Example Using Percent Concentration (m/v) Factors How many mL of a 4.20% (m/v) will contain 3.15 g KCl? STEP 1 Given: 3.15 g of KCl (solute) 4.20% (m/v) KCl solution Need: mL of KCl solution STEP 2 Plan: g of KCl mL of KCl solution 47

Using Percent Concentration(m/v) Factors (continued) STEP 3 Write equalities and conversion factors: 4.20 g KCl = 100 mL of KCl solution 4.20 g KCl and 100 mL KCl solution 100 mL KCl solution 4.20 g KCl STEP 4 Set up the problem: 3.15 g KCl x 100 mL KCl solution = 75.0 mL of KCl 4.20 g KCl

Chapter 8 Solutions 8.5 Molarity and Dilution

Molarity (M) Molarity (M) is a concentration term for solutions gives the moles of solute in 1 L solution molarity (M) = moles of solute liter of solution

Preparing a 1.0 Molar NaCl Solution A 1.0 M NaCl solution is prepared by weighing out 58.5 g of NaCl (1.00 mole) and adding water to make 1.0 liter of solution

Guide to Calculating Molarity

Example of Calculating Molarity What is the molarity of 0.500 L NaOH solution if it contains 6.00 g of NaOH? STEP 1 Given: 6.00 g of NaOH in 0.500 L solution Need: molarity (mole/L) of NaOH solution STEP 2 Plan: g of NaOH moles of NaOH molarity

Example of Calculating Molarity (continued) STEP 3 Write equalitites and conversion factors: 1 mole of NaOH = 40.0 g of NaOH 1 mole NaOH and 40.0 g NaOH 40.0 g NaOH 1 mole NaOH STEP 4 Set up problem to calculate molarity: 6.00 g NaOH x 1 mole NaOH = 0.150 mole of NaOH 40.0 g NaOH 0.150 mole NaOH 0.500 L NaOH solution = 0.300 M NaOH

Molarity Conversion Factors The units of molarity are used to write conversion factors for calculations with solutions.

Example of Calculations Using Molarity How many grams of KCl are needed to prepare 125 mL of a 0.720 M KCl solution? STEP 1 Given: 125 mL (0.125 L) of 0.720 M KCl Need: grams of KCl STEP 2 Plan: L of KCl moles of KCl g of KCl

Example of Calculations Using Molarity (continued) STEP 3 Write equalities and conversion factors: 1 mole of KCl = 74.6 g of KCl 1 mole KCl and 74.6 g KCl 74.6 g KCl 1 mole KCl 1 L of KCl = 0.720 mole of KCl 1 L and 0.720 mole KCl 0.720 mole KCl 1 L STEP 4 Set up problem to cancel mole KCl: 0.125 L x 0.720 mole KCl x 74.6 g KCl = 6.71 g of KCl 1 L 1 mole KCl

Dilution In a dilution, water is added volume increases concentration of solute decreases

Initial and Diluted Solutions In the initial and diluted solution, the moles of solute are the same the concentrations and volumes are related by the following equations: For percent concentration C1V1 = C2V2 Concentrated Diluted solution solution For molarity M1V1 = M2V2

Guide to Calculating Dilution Quantities

Example of Dilution Calculations Using Percent Concentration What volume of a 2.00% (m/v) HCl solution can be prepared by diluting 25.0 mL of 14.0% (m/v) HCl solution? STEP 1 Prepare a table: C1 = 14.0% (m/v) V1 = 25.0 mL C2 = 2.00% (m/v) V2 = ? STEP 2 Solve dilution expression for the unknown C1V1 = C2V2 V2 = V1C1 C2 STEP 3 Set up the problem using known quantities: V2 = V1C1 = (25.0 mL)(14.0%) = 175 mL C2 2.00%

Example of Dilution Calculations Using Molarity What is the molarity (M) of a solution prepared by diluting 0.180 L of 0.600 M HNO3 to 0.540 L? STEP 1 Prepare a table: M1 = 0.600 M V1 = 0.180 L M2 = ? V2 = 0.540 L STEP 2 Solve the dilution expression for the unknown: M1V1 = M2V2 STEP 3 Set up the problem using known quantities: M2 = M1V1 = (0.600 M)(0.180 L) = 0.200 M V2 0.540 L

Molarity in Chemical Reactions In a chemical reaction, the volume and molarity of a solution are used to determine the moles of a reactant or product molarity ( mole ) x volume (L) = moles 1 L if molarity (mole/L) and moles are given, the volume (L) can be determined moles x 1 L = volume (L) moles

Guide to Calculations Involving Solutions in Chemical Reactions

Using Molarity of Reactants How many mL of 3.00 M HCl are needed to completely react with 4.85 g of CaCO3? 2HCl(aq) + CaCO3(s) CaCl2(aq) + CO2(g) + H2O(l) STEP 1 Given: 3.00 M HCl; 4.85 g of CaCO3 Need: volume of HCl in mL STEP 2 Write a plan: g of CaCO3 moles of CaCO3 moles of HCl mL of HCl STEP 3 Write equalities and conversion factors: 1 mole of CaCO3 = 100.1 g 1 mole CaCO3 and 100.1 g CaCO3 100.1 g CaCO3 1 mole CaCO3

Using Molarity of Reactants (continued) STEP 3 (continued) 1 mole of CaCO3 = 2 moles of HCl 1 mole of CaCO3 and 2 moles of HCl 2 moles of HCl 1 mole of CaCO3 1000 mL of HCl = 3.00 moles of HCl 1000 mL of HCl and 3.00 moles of HCl 3.00 moles of HCl 1000 mL of HCl STEP 4 Set up problem to calculate mL of HCl: 4.85 g CaCO3 x 1 mole CaCO3 x 2 moles HCl x 1000 mL HCl 100.1 g CaCO3 1 mole CaCO3 3.00 moles HCl = 32.3 mL of HCl required

8.6 Properties of Solutions Chapter 8 Solutions 8.6 Properties of Solutions

Solutions Solutions contain small particles (ions or molecules) are transparent do not separate, even by filtration or through a semipermeable membrane do not scatter light

Colloids Colloids have medium-sized particles cannot be separated by filtration can be separated by semipermeable membranes scatter light

Examples of Colloids

Suspensions Suspensions have very large particles settle out can be separated by filtration must be stirred to stay suspended Examples: blood platelets, muddy water, and calamine lotion

Solutions, Colloids, and Suspensions

Comparison of Solutions, Colloids, and Suspensions

Colligative Properties are changes in the properties of a solvent when solute particles are added depend on the number of solute particles in solution involve the lowering of freezing point involve an increase in the boiling point

Freezing Point Lowering and Boiling Point Elevation When 1 mole of solute particles is added to 1000 g of water, the freezing point of water decreases by 1.86 °C (from 0 °C to –1.86 °C) boiling point of water increases by 0.52 °C (from 100 °C to 100.52 °C)

Moles of Particles The number of moles of particles depends on the type of solute: nonelectrolytes dissolve as molecules 1 mole of nonelectrolyte = 1 mole of particles in water strong electrolytes dissolve as ions 1 mole of electrolyte = 2 to 4 moles of particles in water

Effect of Solute Particles

Osmosis In osmosis, water (solvent) flows 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

Example of Osmosis A semipermeable membrane separates a 4% starch solution from a 10% starch solution. Starch is a colloid and cannot pass through the membrane, but water can. What happens? semipermeable membrane 4% starch 10% starch H2O

Example of Osmosis (continued) The 10% starch solution is diluted by the flow of water out of the 4% solution, and its volume increases. The 4% solution loses water, and its volume decreases. Eventually, the water flow between the two becomes equal. 7% starch 7% starch H2O

Osmotic Pressure Osmotic pressure is produced by the solute particles dissolved in a solution is the pressure that prevents the flow of additional water into the more concentrated solution increases as the number of dissolved particles in the solution increases

Osmotic Pressure of the Blood Red blood cells have cell walls that are semipermeable membranes maintain an osmotic pressure that cannot change without damage occuring must maintain an equal flow of water between the red blood cell and its surrounding environment

Isotonic Solutions An isotonic solution exerts the same osmotic pressure as red blood cells is known as a “physiological solution” of 5.0% (m/v) glucose or 0.9% (m/v) NaCl is used medically because each has a solute concentration equal to the osmotic pressure equal to red blood cells

Hypotonic Solutions A hypotonic solution has a lower osmotic pressure than red blood cells (RBCs) has a lower concentration than physiological solutions causes water to flow into RBCs causes hemolysis (RBCs swell and may burst)

Hypertonic Solutions A hypertonic solution has a higher osmotic pressure than RBCs has a higher concentration than physiological solutions causes water to flow out of RBCs causes crenation (RBCs shrink in size)

Dialysis In dialysis, solvent and small solute particles pass through an artificial membrane large particles are retained inside waste particles such as urea from blood are removed using hemodialysis (artificial kidney)

Hemodialysis When the kidneys fail, an artificial kidney uses hemodialysis to remove waste particles such as urea from blood.