Part 2: Reactions & Problems involving weak electrolytes Review: Strong Electrolytes
Remember: Strong electrolytes ionize or dissociate completely. ______electrolytes only dissociate to a small extent. To solve problems that involve weak acids and weak bases, we will revisit Equilibrium.
Three classes of strong electrolytes 1. ______2. ______3. ______
2 Review: Strong Electrolytes Calculation of concentrations of ions in solution of strong electrolytes is fairly easy
~ Calculate the concentrations of ions in 0.070 M
Not as simplistic with weak acids and weak bases. 3 ACID-BASE THEORIES ______ acid--donates a hydrogen ion (H+) in water base--donates a hydroxide ion in water (OH−) This theory iswas limited
______ acid--donates a proton in water base--accepts a proton in water This theory is better; it explains ammonia as a base! This is the main theory that we will use for our acid/base discussion.
______ acid--accepts an electron pair base--donates an electron pair This theory explains most traditional acids and bases plus a host of coordination compounds and is used widely in organic chemistry. Uses coordinate covalent bonds. ACIDS DONATE ONLY ONE PROTON AT A TIME
______—acids that can donate one H+ ex.
______—acids that can donate two H+'s ex.
______—acids that can donate many H+'s ex. Acid Dissociation (Ionization, Protonation) Reactions Ex. 1) Write the simple dissociation (ionization) reaction (omitting water) for each of the following acids. a. Hydrochloric acid (HCl)
b. Acetic acid (CH3COOH or HC2H3O2)
+ c. The ammonium ion (NH4 )
+ d. The anilinium ion (C6H5NH3 ) The water concentration in dilute aqueous solutions is very high. 1 L of water contains 55.5 moles of water. Thus in dilute aqueous solutions:
H 2 O 55 .5 M • The water concentration is many orders of magnitude greater than the ion concentrations. • Thus the water concentration is essentially ______. Which is why we can omit water from our equilibrium expressions. 7 The Auto-Ionization of Water Pure water ionizes ______ less than one-millionth molar + - H2O + H2O H3O + OH Because the activity of pure water is 1, the equilibrium constant for this reaction is
+ Kc H3O OH
8 Experimental measurements have determined that the concentration of each ion is 1.0 x 10-7 M at 250C + Kc H3O OH 1.0 x 10-7 1.0 x 10-7 1.0 x1014 This particular equilibrium constant is called the ion-product for water, Kw so Kw = ______9 The pH and pOH scales Remember: A convenient way to express ______and ______is through pH. pH is defined as + pH = -logH3O
+ - If we know either [H3O ] or [OH ], then we can calculate ______and vise versa. + ^-pH - ^-pOH [H3O ] = 10 [OH ] = 10 14 H3O OH 1.010 pH pOH 14.00 10 + - Ex. 2) Calculate the number of H3O and OH ions in one liter of pure water at 250C.
11 pH Scale
Notice: When [H+] = 10-3 pH = 3
When [H+] = 10-9 pH = 9 Develop 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. + - [H3O ] [OH ] pH pOH 1.0 M 1.0 x 10-14 M 0.00 14.00 1.0 x 10-3 M 1.0 x 10-11 M 3.00 11.00 1.0 x 10-7 M 1.0 x 10-7 M 7.00 7.00 2.0 x 10-12 M 5.0 x 10-3 M 11.70 2.30 1.0 x 10-14 M 1.0 M 14.00 0.00
13 WEAK ACIDS AND BASES: The vast majority of acids/bases are weak. They are considered weak b/c they do not ionize very much.
This means an equilibrium is established and it lies far to the left (reactant favored).
The equilibrium expression for acids is known as the Ka (the acid dissociation constant). It is set up the same way as any other equilibrium expression. Ionization Constants for Weak Monoprotic Acids and Bases Let’s look at the dissolution of acetic acid, a weak acid, in water as an example. The equation for the ionization of acetic acid is: - CH3COOH H2O H3O CH3COO The equilibrium constant for this ionization is expressed as:
15 Values for several ionization constants ACID FORMULA IONIZATION CONSTANT -5 Acetic CH3COOH 1.8 x 10 -4 Nitrous HNO2 5.1 x 10 Hydrofluoric HF 7.2 x 10-4 Hypochlorous HClO 2.9 x 10-8 Hydrocyanic HCN 6.0 x 10-10
Other values are found on Table B.8
16 From 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:
17 Ex. 3) Write the equation for the ionization of the weak
acid HCN and the expression, Ka for its ionization constant. Also label the conjugate acid/base pairs
18 Calculations involving Weak Acids
Use the RICE TABLE method you learned in general equilibrium
Set up an equilibrium reaction. Start by writing the balanced equation Define initial concentrations, changes, and final concentrations in terms of x
Set up the acid equilibrium expression (Ka) and solve for Ka OR
Substitute values and variables into the Ka expression and solve for x. Calculation of Ionization Constants Ex. 4) In 0.12 M solution, a weak monoprotic acid, HX, 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.
20 Ex. 5) 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-pH
21 Calculations Based on Ionization Constants Ex. 6) Calculate the concentrations of the various species in 0.15 M acetic acid, CH3COOH, solution. -5 Ka = 1.8 x 10 Always write down the ionization reaction and the ionization constant expression. - CH3COOH H2O H3O CH3COO - H3O CH3COO 5 Ka 1.810 CH3COOH
22 Reminder: Combine the basic chemical concepts with some algebra to solve the problem
Substitute the algebraic quantities into the ionization expression.
-4 Solve the algebraic equation. If Ka is less than x 10 you can cancel out – x or + x , using the simplifying assumption (happy face rule)
Complete the algebra and solve for concentrations. 23 Note that the properly applied simplifying assumption gives the same result as solving the quadratic equation does. x x 1.8 10 5 0.15 X x 2 1.8 10 5 x 2.7 10 6 0 a b c b b 2 4ac x 2a
2 1.8105 1.8105 41 2.7106 x 21 x 1.6103 and -1.610-3 24 Ex. 7) Now calculate the percent ionization for the 0.15 M acetic acid. From Ex. 6, we know the concentration -3 of CH3COOH that ionizes in this solution is 1.6 x 10 M. The percent ionization of acetic acid is
% ionization = [CH3COOH] ionized x 100% [CH3COOH] original
25 Ex. 8) Calculate the concentrations of the species in 0.15 M hydrocyanic acid, HCN, solution. Then find the % -10 ionization. Ka=4.0 x 10 for HCN
26 Let’s look at the percent ionization of two weak acids as a function of their ionization constants for Ex. 7 & 8
The [H+] in 0.15 M acetic acid is ______times greater than for 0.15 M HCN.
+ Solution Ka [H ] pH % ionization 0.15 M 1.8 x 10-5 1.6 x 10-3 2.80 1.1 CH3COOH 0.15 M 4.0 x 10-10 7.7 x 10-6 5.11 0.0051 HCN
27 Calculations involving Weak Bases Weak bases work the same way as weak acids. However, the x you solve for is the [OH−] and taking the negative log of x will give you the pOH and not the pH!
Ex. 9) Calculate the concentrations of the various species in 0.25 M aqueous ammonia and the -5 percent ionization. Kb for ammonia = 1.8 x 10
28 Calculations Based on Ionization Constants Ex. 10) 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.
29 Determination of the pH of a Mixture of Weak Acids
Only the acid with the largest Ka value will contribute an appreciable [H+]. Determine the pH based on this acid and ignore any others. Determination of the pH of a Mixture of Weak Acids Ex. 11) a. Calculate the pH of a solution that contains 5.00 M −10 −4 HCN (Ka = 6.2 × 10 ) and 5.00 M HNO2 (Ka = 4.0 × 10 ). b. Calculate the concentration of cyanide ion (CN−) in this solution at equilibrium. Polyprotic Acids Many weak acids contain two or more acidic hydrogens.
Acids with more than one ionizable hydrogen Ionize in steps
Each dissociation has its own Ka value.
The first dissociation will be the greatest and subsequent dissociations will have much smaller Ka As each H+ is removed, the remaining acid gets weaker and therefore has a smaller equilibrium constants. As the negative charge on the acid increases it becomes more difficult to remove the positively charged proton. Polyprotic Acids
Consider arsenic acid, H3AsO4, which has three ionization constants -4 1 K1=2.5 x 10 -8 2 K2=5.6 x 10 -13 3 K3=3.0 x 10
33 Polyprotic Acids The first ionization step is
34 Polyprotic Acids The second ionization step is
35 Polyprotic Acids The third ionization step is
36 Polyprotic Acids Notice that the ionization constants vary in the following fashion: K1 K2 K3 This is a general relationship.
37 Polyprotic Acids Ex. 12) 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. Can’t use assumption. (too close for the st + 1 H ) 38 4. 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. 39 A comparison of the various species in 0.100 M H3AsO4 solution follows. Species Concentration H3AsO4 0.095 M H+ 0.0049 M - H2AsO4 0.0049 M 2- -8 HAsO4 5.6 x 10 M 3- -18 AsO4 3.4 x 10 M OH- 2.0 x 10-12 M
40 Ka and pKa
pKa = - log Ka
The stronger the acid, the higher the Ka value and the lower the pKa value Ka and pKa Ex. 13) The pH of 0.10 M HClO solution was measured to
be 4.77 What is the value of the pKa for HClO? Solvolysis ______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.
43 Hydrolysis at 25oC + - -7 in ______solutions: [H3O ] = [OH ] = 1.0 x 10 M + - -7 in ______solutions:[H3O ] < [OH ] > 1.0 x 10 M - + -7 in ______solutions:[OH ] < [H3O ] > 1.0 x 10 M
for all conjugate acid/base pairs in aqueous solns.
Kw = Ka Kb
So if we know the value of either Ka or Kb, the other can be calculated. 44 Salts of Strong Soluble Bases and Weak Acids A H2O HA OH HA OH KW Kb = A Ka for HA
Note: This same method can be applied to the anion of any weak monoprotic acid. 45 Salts of Strong Soluble Bases and Weak Acids Ex. 14) Calculate the hydrolysis constants for the following anions of weak acids. a) F- , fluoride ion, the anion of hydrofluoric acid, HF. -4 For HF Ka= 7.2 x 10 .
46 Salts of Strong Soluble Bases and Weak Acids Ex. 14) Calculate the hydrolysis constants for the following anions of weak acids. b) CN-, cyanide ion, the anion of hydrocyanic acid, -10 HCN. For HCN, Ka= 4.0 x 10 .
47 Ex. 15) 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.
~100%inH O NaClO 2 Na ClO 0.10M 0.10M 0.10M
48 Salts of Acids and Bases Aqueous 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 ______.
49 A Fun Chemistry Problem for you
Rain 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 -7 earth? Ka for H2 CO3 = 4.2 x 10
50