 # Properties of Solutions. Classification of Matter Solutions are homogeneous mixtures.

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Properties of Solutions

Classification of Matter Solutions are homogeneous mixtures

A. Concentration The amount of solute in a solution. Describing Concentration –% by mass - medicated creams –% by volume - rubbing alcohol –ppm, ppb - water contaminants –molarity - used by chemists –molality - used by chemists

Solute A solute is the dissolved substance in a solution. A solvent is the dissolving medium in a solution. Solvent Salt in salt water Sugar in soda drinks Carbon dioxide in soda drinks Water in salt waterWater in soda

MOLARITY

A unit on concentration that is the ratio between moles of DISSOLVED substance and liters of solution

Molarity (M) = moles of solute (mol) volume of solution (L)

There are many ways to represent molarity: Molarity M Molar mol / L

MAKE SURE YOUR UNITS ARE CORRECT!!! ALWAYS mol/L

MOLARITY A measurement of the concentration of a solution Molarity (M) is equal to the moles of solute (n) per liter of solution M = n / V = mol / L Calculate the molarity of a solution prepared by mixing 1.5 g of NaCl in 500.0 mL of water. First calculate the moles of solute: 1.5 g NaCl ( 1 mole NaCl ) = 0.0257 moles of NaCl 58.45 g NaCl Next convert mL to L: 0.500 L of solution Last, plug the appropriate values into the correct variables in the equation: M = n / V = 0.0257 moles / 0.500 L = 0.051 mol/L

MOLARITY M = n / V = mol / L How many grams of LiOH is needed to prepare 250.0 mL of a 1.25 M solution? First calculate the moles of solute needed: M = n / V, now rearrange to solve for n: n = MV n = (1.25 mol / L ) (0.2500 L) = 0.3125 moles of solute needed Next calculate the molar mass of LiOH: 23.95 g/mol Last, use deminsional analysis to solve for mass: 0.3125 moles (23.95 g LiOH / 1 mol LiOH) = 7.48 g of LiOH

Problem 4:67 pg 153 a,b

C. Dilution Preparation of a desired solution by adding water to a concentrate. Moles of solute remain the same.

C. Dilution What volume of 15.8M HNO 3 is required to make 250 mL of a 6.0M solution? GIVEN: M 1 = 15.8M V 1 = ? M 2 = 6.0M V 2 = 250 mL WORK: M 1 V 1 = M 2 V 2 (15.8M) V 1 = (6.0M)(250mL) V 1 = 95 mL of 15.8M HNO 3

Calculate the molarity of a solution prepared by diluting 25.0 mL of 0.05 M potassium iodide with 50.0 mL of water (the densities are similar). M 1 = 0.05 mol/LM 2 = ? V 1 = 25.0 mLV 2 = 50.0 + 25.0 = 75.0 mL M 1 V 1 = M 2 V 2 M 1 V 1 = M 2 = (0.05 mol/L) (25.0 mL) = 0.0167 M of KI V 2 75.0 mL MOLARITY & Dilution

Problem 4:75 pg 153 a,b

Chapter 4 Aqueous Reactions and Solution Stoichiometry

Electrolytes Substances that dissociate into ions when dissolved in water. A nonelectrolyte may dissolve in water, but it does not dissociate into ions when it does so.

Electrolytes and Nonelectrolytes Soluble ionic compounds and strong acids tend to be electrolytes. There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4

Electrolytes and Nonelectrolytes Molecular compounds tend to be nonelectrolytes, except for acids and bases.

Electrolytes A strong electrolyte dissociates completely when dissolved in water. A weak electrolyte only dissociates partially when dissolved in water.

Strong Electrolytes Are… Strong acids Strong bases Soluble ionic salts

1.Pure water 2.Tap water 3.Sugar solution 4.Sodium chloride solution 5.Hydrochloric acid solution 6.Lactic acid solution 7.Ethyl alcohol solution 8.Pure sodium chloride Electrolytes?

ELECTROLYTES: NONELECTROLYTES: Tap water (weak) NaCl solution HCl solution Lactate solution (weak) Pure water Sugar solution Ethanol solution Pure NaCl Answers to Electrolytes

Problem 4:15 pg 150

Precipitation Reactions When one mixes ions that form compounds that are insoluble (as could be predicted by the solubility guidelines), a precipitate is formed.

Metathesis (Exchange) Reactions Double replacement Metathesis comes from a Greek word that means “to transpose” It appears the ions in the reactant compounds exchange, or transpose, ions AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq)

Net Ionic Equation Need to know which ionic bonds are soluble and which form precipitates Need to use solubility chart

Writing Net Ionic Equations 1.Write a balanced molecular equation. 2.Dissociate all strong electrolytes. 3.Cross out anything that remains unchanged from the left side to the right side of the equation. 4.Write the net ionic equation with the species that remain.

Net Ionic Equation Those things that didn’t change (and were deleted from the net ionic equation) are called spectator ions. Ag + (aq) + NO 3 - (aq) + K + (aq) + Cl - (aq)  AgCl (s) + K + (aq) + NO 3 - (aq) AgNO 3 (aq) + KCl (aq)  AgCl (s) + KNO 3 (aq) Not soluble

Net Ionic Equation Ag + (aq) + Cl - (aq)  AgCl (s)

Problem 4.19pg 151 4.39 b and C

Acids There are only seven strong acids: Hydrochloric (HCl) Hydrobromic (HBr) Hydroiodic (HI) Nitric (HNO 3 ) Sulfuric (H 2 SO 4 ) Chloric (HClO 3 ) Perchloric (HClO 4 )

Bases The strong bases are the soluble salts of hydroxide ion (OH - ): Alkali metals Calcium Strontium Barium

Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… HCl (aq) + NaOH (aq)  NaCl (aq) + H 2 O (l) H + ( aq ) + Cl - ( aq ) + Na + ( aq ) + OH - ( aq )  Na + ( aq ) + Cl - ( aq ) + H 2 O ( l ) H + ( aq ) + Cl - (aq) + Na + ( aq ) + OH - ( aq )  Na + ( aq ) + Cl - ( aq ) + H 2 O ( l )

Neutralization Reactions When a strong acid reacts with a strong base, the net ionic equation is… H + ( aq )+ OH - ( aq )  H 2 O ( l )

Neutralization Reactions Observe the reaction between Milk of Magnesia, Mg(OH) 2, and HCl.

Problem 4.81g 153

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