Acid and Base Equilibrium

Acid and Base Equilibrium Chapter 182Strong ElectrolytesStrong electrolytes _________ or _________ completely

Three classes of strong electrolytes1. __________________ 2. _________________ 3. __________________

23Strong ElectrolytesCalculation of concentrations of ions in solution of strong electrolytes is fairly easyEx.1) Calculate the concentrations of ions in 0.050 M nitric acid, HNO3.

34Strong ElectrolytesEx. 2) Calculate the concentrations of ions in 0.050 M calcium hydroxide, Ca(OH)2, solution.

45The water concentration in dilute aqueous solutions is very high.1 L of water contains 55.5 moles of water.Thus in dilute aqueous solutions: The water concentration is many orders of magnitude greater than the ion concentrations. Thus the water concentration is essentially constant.56The Auto-Ionization of WaterPure water ionizes _________ less than one-millionth molar

Because the activity of pure water is 1, the equilibrium constant for this reaction is

67Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 250C

This particular equilibrium constant is called the _________ ____________________________________ 78A convenient way to express _________ and _________ is through pH. pH is defined as

If we know either [H3O+] or [OH-], then we can calculate _________________ and vise versa.[H3O+] = 10^-pH[OH-] = 10^-pOH

The pH and pOH scales89Ex. 3) Calculate the concentrations of H3O+ and OH- in 0.050 M HCl and find the pH of the solution.

910Ex. 4) The pH of a solution is 4.597. What is the concentration of H3O+?

1011Ex. 5) Calculate [H3O+], pH, [OH-], and pOH for 0.020 M Ba(OH)2 solution.

1112Ex. 6) Calculate the number of H3O+ and OH- ions in one liter of pure water at 250C.

1213Develop familiarity with pH scale by looking at a series of solutions in which [H3O+] varies between 1.0 M and 1.0 x 10-14 M.

1314Ionization Constants for Weak Monoprotic Acids and BasesLets look at the dissolution of acetic acid, a weak acid, in water as an example.The equation for the ionization of acetic acid is:

The equilibrium constant for this ionization is expressed as:

1415Ionization Constants for Weak Monoprotic Acids and BasesValues for several ionization constants

1516Ionization Constants for Weak Monoprotic Acids and BasesFrom the above table we see that the order of increasing acid strength for these weak acids is:

The order of increasing base strength of the anions (conjugate bases) of these acids is:

1617Ionization Constants for Weak Monoprotic Acids and BasesEx. 7) Write the equation for the ionization of the weak acid HCN and the expression for its ionization constant.1718Calculation of Ionization ConstantsEx. 8) In 0.12 M solution, a weak monoprotic acid, HY, is 5.0% ionized. a) Calculate the concentrations of all species in solution.b) Calculate the ionization constant for the weak acid.

Since the weak acid is 5.0% ionized, it is also 95% unionized.

1819Calculation of Ionization ConstantsEx. 9) The pH of a 0.100 M solution of a weak monoprotic acid, HA, is found to be 2.970. What is the value for its ionization constant?pH = 2.970 so [H+]= 10-pH1920Calculations Based on Ionization ConstantsEx. 10) Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. Ka = 1.8 x 10-5Always write down the ionization reaction and the ionization constant expression.

2021Calculations Based on Ionization ConstantsFollow these steps

Combine the basic chemical concepts with some algebra to solve the problemSubstitute the algebraic quantities into the ionization expression.Solve the algebraic equation. If Ka is less than x 10-4 you can cancel out x or + x , using the simplifying assumptionComplete the algebra and solve for concentrations.

2122Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

2223Calculations Based on Ionization ConstantsEx. 11) Now calculate the percent ionization for the 0.15 M acetic acid. From Ex. 10, we know the concentration of CH3COOH that ionizes in this solution is 1.6 x 10-3 M. The percent ionization of acetic acid is

% ionization = [CH3COOH] ionized x 100% [CH3COOH] original 2324Calculations Based on Ionization ConstantsEx. 12) Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Then find the % ionization.Ka=4.0 x 10-10 for HCN2425Lets look at the percent ionization of two weak acids as a function of their ionization constants for Ex. 11 & 12

The [H+] in 0.15 M acetic acid is _________ times greater than for 0.15 M HCN.

2526Weak bases work the same way as weak acids

Ex. 13) Calculate the concentrations of the various species in 0.25 M aqueous ammonia and the percent ionization. Kb for ammonia = 1.8 x 10-5

2627Calculations Based on Ionization ConstantsEx. 14) The pH of an aqueous ammonia solution is 11.37. Calculate the molarity (original concentration) of the aqueous ammonia solution.Use the ionization expression and some algebra to get the equilibrium concentration.

2728Polyprotic AcidsMany weak acids contain two or more acidic hydrogens.polyprotic acids ionize stepwiseionization constant for each stepConsider arsenic acid, H3AsO4, which has three ionization constantsK1=2.5 x 10-4K2=5.6 x 10-8K3=3.0 x 10-132829Polyprotic AcidsThe first ionization step is2930Polyprotic AcidsThe second ionization step is3031Polyprotic AcidsThe third ionization step is3132Polyprotic AcidsNotice that the ionization constants vary in the following fashion:

This is a general relationship.3233Polyprotic AcidsEx. 15) Calculate the concentration of all species in 0.100 M arsenic acid, H3AsO4, solution.

1. Write the first ionization ionization step and represent the concentrations.

2. Substitute into the expression for K1.

3. Use the quadratic equation to solve for x, and obtain two values. Cant use assumption. (too close)

33344. ionization and represent the concentrations.

5. Substitute into the second step ionization expression.

6. Now we repeat the procedure for the third ionization step.

7. Substitute into the third ionization expression.

8. Last Use Kw to calculate the [OH-] in the 0.100 M H3AsO4 solution.

3435Polyprotic AcidsA comparison of the various species in 0.100 M H3AsO4 solution follows.

3536Solvolysis_________ is the reaction of a substance with the solvent in which it is dissolved.

_________ refers to the reaction of a substance with water or its ions.

Combination of the anion of a weak acid with H3O+ ions from water to form nonionized weak acid molecules.

3637Hydrolysisat 25oCin _______solutions: [H3O+] = [OH-] = 1.0 x 10-7 Min _______ solutions:[H3O+] < [OH-] > 1.0 x 10-7 Min _______ solutions:[OH-] < [H3O+] > 1.0 x 10-7 M

for all conjugate acid/base pairs in aqueous solns. So if we know the value of either Ka or Kb, the other can be calculated.3738Salts of Strong Soluble Bases and Weak Acids

Note: This same method can be applied to the anion of any weak monoprotic acid.3839Salts of Strong Soluble Bases and Weak AcidsEx. 16) Calculate the hydrolysis constants for the following anions of weak acids. a) F-, fluoride ion, the anion of hydrofluoric acid, HF. For HF, Ka=7.2 x 10-4. b) CN-, cyanide ion, the anion of hydrocyanic acid, HCN. For HCN, Ka= 4.0 x 10-10.3940Salts of Strong Soluble Bases and Weak AcidsEx. 17) Calculate [OH-], pH and percent hydrolysis for the hypochlorite ion in 0.10 M sodium hypochlorite, NaClO, solution. Clorox, Purex, etc., are 5% sodium hypochlorite solutions.

4041Salts of Acids and BasesAqueous solutions of salts of strong acids and strong soluble bases are _______.Aqueous solutions of salts of strong bases and weak acids are _______. Aqueous solutions of salts of weak bases and strong acids are _______.Aqueous solutions of salts of weak bases and weak acids can be _______ _______ _______ __.4142Rain water is slightly acidic because it absorbs carbon dioxide from the atmosphere as it falls from the clouds. (Acid rain is even more acidic because it absorbs acidic anhydride pollutants like NO2 and SO3 as it falls to earth.) If the pH of a stream is 6.5 and all of the acidity comes from CO2, how many CO2 molecules did a drop of rain having a diameter of 6.0 mm absorb in its fall to earth? Ka for H2 CO3 = 4.2 x 10-7 42[H3O+][OH-]pHpOH

1.0 M1.0 x 10-14 M0.0014.00

1.0 x 10-3 M1.0 x 10-11 M3.0011.00

1.0 x 10-7 M1.0 x 10-7 M7.007.00

2.0 x 10-12 M5.0 x 10-3 M11.702.30

1.0 x 10-14 M1.0 M14.000.00

ACIDFORMULAIONIZATION CONSTANT

AceticCH3COOH1.8 x 10-5

NitrousHNO24.5 x 10-4

HydrofluoricHF7.2 x 10-4

HypochlorousHClO3.5 x 10-8

HydrocyanicHCN4.0 x 10-10

SolutionKa[H+]pH% ionization

0.15 M CH3COOH1.8 x 10-51.6 x 10-32.801.1

0.15 M

HCN4.0 x 10-107.7 x 10-65.110.0051

SpeciesConcentration

H3AsO40.095 M

H+0.0049 M

H2AsO4-0.0049 M

HAsO42-5.6 x 10-8 M

AsO43-3.4 x 10-18 M

OH-2.0 x 10-12 M