Acid and Base Equilibrium

Acid and Base Equilibrium Chapter 182Strong ElectrolytesStrong electrolytes ionize or dissociate completely

Three classes of strong electrolytes1. Strong Acids2. Strong Soluble Bases3. Most Soluble Salts

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.

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

34The 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.45The Auto-Ionization of WaterPure water ionizes very slightlyless than one-millionth molar

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

56Experimental 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 ion-product for water, Kw so Kw = 1.0 x 10-14

67A convenient way to express acidity and basicity is through pH. pH is defined as

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

The pH and pOH scales

78Ex. 3) Calculate the concentrations of H3O+ and OH- in 0.050 M HCl and find the pH of the solution.

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

89Ex. 5) Calculate [H3O+], pH, [OH-], and pOH for 0.020 M Ba(OH)2 solution.

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

910Develop 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.

1011Ionization 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:

1112Ionization Constants for Weak Monoprotic Acids and BasesValues for several ionization constants

1213Ionization Constants for Weak Monoprotic Acids and BasesFrom the above table we see that the order of increasing acid strength for these weak acids is:

HF > HNO2 > CH3COOH >HClO >HCN

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

F- < NO2- < CH3COO- < ClO- < CN-1314Ionization 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.1415Calculation 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.

1516Calculation 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-pH1617Calculations 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.

1718Calculations 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.

1819Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does.

1920Calculations 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 2021Calculations 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 HCN2122Lets 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 210 times greater than for 0.15 M HCN.

2223Weak 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

2324Calculations 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.

2425Polyprotic 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 10-132526Polyprotic AcidsThe first ionization step is2627Polyprotic AcidsThe second ionization step is2728Polyprotic AcidsThe third ionization step is2829Polyprotic AcidsNotice that the ionization constants vary in the following fashion:

This is a general relationship.

2930Polyprotic 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)

30314. 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.

3132Polyprotic AcidsA comparison of the various species in 0.100 M H3AsO4 solution follows.

3233SolvolysisSolvolysis is the reaction of a substance with the solvent in which it is dissolved.

Hydrolysis 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.

3334Hydrolysisat 25oCin neutral solutions: [H3O+] = [OH-] = 1.0 x 10-7 Min basic solutions:[H3O+] < [OH-] > 1.0 x 10-7 Min acidic solutions:[OH-] < [H3O+] > 1.0 x 10-7 M

for all conjugate acid/base pairs in aqueous solns. Kw = Ka KbSo if we know the value of either Ka or Kb, the other can be calculated.3435Salts of Strong Soluble Bases and Weak Acids

Note: This same method can be applied to the anion of any weak monoprotic acid.3536Salts 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 10-10.3637Salts 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.

3738Salts of Acids and BasesAqueous solutions of salts of strong acids and strong soluble bases are neutral.Aqueous solutions of salts of strong bases and weak acids are basic. Aqueous solutions of salts of weak bases and strong acids are acidic.Aqueous solutions of salts of weak bases and weak acids can be neutral, basic or acidic.3839Rain 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 39[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