Acids, Bases, and Salts Module 6: Characteristics of Acids and Bases Strong and Weak Acids and Bases Ionization of Water pH and pOH Neutralization Reactions Buffers How Buffers Maintain pH Constant Acidosis and Alkalosis

Properties of Acids  Produce hydrogen (H+) or hydronium (H3O+) ions when dissolved in water  Taste: Sour  Are electrolytes. (Electrolytes are substances that conduct electricity.)  Turns blue litmus paper red  Neutralizes bases  Has a pH level ranging from -1 to 7

Properties of Bases  Produce OH- in aqueous solutions.  Taste: Bitter  Turn red litmus paper blue  Are electrolytes.  e)React with acids.  f)Feel slippery or soapy  Has a pH level ranging from 7 to 14

Acids and Bases H2OH2O Acid Base

Strong and Weak Acids and Bases Strong acids and bases dissociate completely, so they are considered strong electrolytes Compounds that dissociates less than 50% are considered weak electrolytes (weak acids and bases).

Strong Acids  Strong Acid: are strong electrolytes.  Strong acids ionize completely.  A strong acid examples are: HCl, HBr, HI, HClO 4, HNO 3, H 2 SO 4

Strong Bases  Strong Base: are also strong electrolytes  All salts/bases dissociate 100%  A strong base consists of: a [Group 1A metal] + [Hydroxides] and Ca(OH) 2, Sr(OH) 2, Ba(OH) 2

Acids and Bases SubstanceStrong ElectrolyteWeak Electrolyte HBr Ni(OH) 2 KOH HCN Tap water

Acids and Bases SubstanceStrong ElectrolyteWeak Electrolyte HBr  Ni(OH) 2  KOH  HCN  Tap water 

Ionization of Water The above equation is often simplified as follows: H + + OH - H 2 O H 2 O + H 2 O H 3 O + + OH - hydronium hydroxide The double arrow indicates that this is an equilibrium

pH and pOH Notice that pH + pOH = neutral pH pOH [H + ] [OH - ]

pH and pOH pHpOH[H+][OH-] acidic, basic, or neutral

pH and pOH pHpOH[H+][OH-] acidic, basic, or neutral acidic basic neutral basic

Neutralization Reactions

NiCl 3 + LiOH  Ni(OH) 3 + LiCl

Neutralization Reactions NiCl 3 + LiOH  Ni(OH) 3 + LiCl NiCl 3 + LiOH  Ni(OH) LiCl

Neutralization Reactions NiCl 3 + LiOH  Ni(OH) 3 + LiCl NiCl 3 + LiOH  Ni(OH) LiCl NiCl LiOH  Ni(OH) LiCl

Buffers A buffer solution is a solution whose pH remains relatively constant when either acids or bases are added to it. Buffers do not have an unlimited ability to resist pH changes. This is the buffering capacity.

Buffers A buffer solution is a solution whose pH remains relatively constant when either acids or bases are added to it. Buffers do not have an unlimited ability to resist pH changes. This is the buffering capacity.

Buffers Buffers must be able to react with excess acid or excess base It has two components one to react with the acid and one to react with the base Most common buffers consist of: a weak acid and its salt or a weak base and its salt

Example Calculate the mass of NaCl found in 200 mL of 6.0% (m/v) NaCl solution.

Example What volume of a solution that is 5.0% (m/v) copper (II) nitrate contains 50.0g of Cu(NO 3 ) 2 ?

Example What volume of a solution that is 5.0% copper (II) nitrate contains 50.0g of Cu(NO 3 ) 2 ?

Example What volume of a solution that is 5.0% copper (II) nitrate contains 50.0g of Cu(NO 3 ) 2 ?

Example What volume of a solution that is 5.0% copper (II) nitrate contains 50.0g of Cu(NO 3 ) 2 ?

Example Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive).

Example Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive). First, we need to determine the total volume of the solution: 25mL ethanol + 50 mL water = 75 mL solution

Example Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive). First, we need to determine the total volume of the solution: 25mL ethanol + 50 mL water = 75 mL solution

Example Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive). First, we need to determine the total volume of the solution: 25mL ethanol + 50 mL water = 75 mL solution

Example Calculate the percent v/v of a solution prepared by dissolving 25 ml of ethanol in 50mL water (assume the volumes are additive). First, we need to determine the total volume of the solution: 25mL ethanol + 50 mL water = 75 mL solution

Molarity Moles solute = M Liter of solution

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example What is the molarity (M) of a solution that contains 30.0g copper (II) nitrate in 250mL solution?

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Example Calculate the mass of NaOH needed to prepare 150mL of a 0.220M solution of sodium hydroxide.

Dilutions V 1 X C 1 = V 2 X C 2 where C is concentration, it may be Molarity, mass%, or other concentration units

Example Calculate the volume of 6.0MKCl needed to prepare 250mL of a 0.300M solution of potassium chloride.

Example Calculate the volume of 6.0MKCl needed to prepare 250mL of a 0.300M solution of potassium chloride. V 1 X C 1 = V 2 X C 2

Example Calculate the volume of 6.0MKCl needed to prepare 250mL of a 0.300M solution of potassium chloride. V 1 X C 1 = V 2 X C 2

Example Calculate the volume of 6.0MKCl needed to prepare 250mL of a 0.300M solution of potassium chloride. V 1 X C 1 = V 2 X C 2

Solutions, Colloids and Suspensions SOLUTIONCOLLOIDSSUSPENSIONS Solute Particles:small particles (ions or molecules) larger particlesvery large particles (macromolecules) Examples:salted water, beerstarch and water, fog, smoke, sprays calamine lotion Characteristics: Transparent. Particles cannot be separated by filters or semipermeable membranes. Tyndall Effect. Particles can be separated by semi-permeable membranes. Opaque. Particles can be separated by filters.

Osmosis The flow of a solvent through a semipermeable membrane into a solution of higher solute concentration. A semipermeable membrane, such as cellophane, contains tiny holes far too small to be seen but large enough to let solvent molecules pass through, but NOT large solute particles.

Osmosis Compartment A – pure water Compartment B – glucose solution dissolved AB

Osmosis dissolved AB osmotic pressure Osmosis reaches an equilibrium and the level of compartment B is higher than A. These levels can be made equal again by applying an external pressure to compartment B. The amount of pressure necessary to equalize both levels is called osmotic pressure.

Osmotic Pressure Depends on the number of particles in the solution. In the blood, SolutionOsmotic pressure Examples hypotonic Lower than red blood cells Pure water, solutions LESS than 5% glucose, 0.9% NaCl solution isotonic Same as red blood cells Plasma, 5% glucose, 0.9% NaCl solution hypertonic Higher than red blood cells Solutions GREATER than 5% glucose, 0.9% NaCl solution

Dialysis Uses a semipermeable membrane to separate dissolved ions and molecules from larger colloidal particles dispersed in the solution. In kidneys the waste products of the blood dialyze out through a semipermeable membrane.

Acid, Bases and Salts THE END