equilibration of the reaction product. ⢠The key idea is to let what are combined react 100% as a limiting reagent problem as a preliminary step done before.
INTRODUCTION (1) • On first encounter, the study of acidbase equilibria is a little like a strange land with seemingly confusing trails that make passage difficult. • In fact, there is a road map that, once understood, allows us to navigate acid-base equilibria with confident precision and so become masters of its domain. • Here is an overview of this road map.
INTRODUCTION (2) • Fundamentally, acid-base equilibria are just a particular example of the ideas and techniques we have already learned in the study of gas phase chemical equilibria. • However, there are two aspects that complicate the application of these ideas. • First, because the auto-ionization of water is always present in aqueous solution, the analysis of aqueous acid-base equilibria must always take into account at least two competing equilibria, the acid or base ionization and the water auto-ionization.
INTRODUCTION (3) • Second, because we will be interested in how acid-base equilibria respond to changes in the system (typically by adding additional base or acid). • We need also to be able to separate the chemical reactions that take place when things are combined from the subsequent equilibration of the reaction product. • The key idea is to let what are combined react 100% as a limiting reagent problem as a preliminary step done before equilibration.
INTRODUCTION (4) • Our overarching goal, then, is to learn to clearly distinguish and to separately master these two aspects: Competing equilibria, and Limiting reagent reactions. • Once this is achieved, we will have a framework in which any problem in acid-base equilibria can be solved in a straightforward way.
OBJECTIVES To familiarize with water autoionization, pH, pOH and pKa terms. To become familiar with dissociation constants determination; To introduce acid and bases strength
TEXTBOOK Brown, Lemay & Bursten, Chemistry: The Central Science, 10th Ed. (Chapter 16)
AUTOIONIZATION OF WATER • As we have seen, water is amphoteric. • In pure water, a few molecules act as bases and a few act as acids. H2O(l) + H2O(l) D H3O+(aq) + OH−(aq) • This is referred to as auto-ionization. • The equilibrium expression for this process is: Kc = [H3O+] [OH−] • This special equilibrium constant is referred to as the ion-product constant for water (Kw). • At 25ºC: Kw = 1.0*10−14
pH (1) • pH is defined as the negative base -10 logarithm of the hydronium ion concentration. pH = −log [H3O+] • In pure water: Kw = [H3O+]*[OH−] = 1.0*10−14 • Because in pure water: [H3O+] = [OH−] • Then: [H3O+] = (1.0*10−14)1/2 = 1.0*10−7 • Therefore, for pure water: pH = −log (1.0*10−7) = 7.00
pH (2) • An acid has a higher [H3O+] than pure water, so its pH is < 7. • A base has a lower [H3O+] than pure water, so its pH is > 7.
pH (3) These are the pH values for several common substances.
Other “p” Scales • The “p” in pH tells us to take the negative log of the quantity (in this case, hydrogen ions). • Some similar examples are: pOH = −log [OH−] pKw = −log Kw
WATCH THIS! Because: [H3O+]*[OH−] = Kw = 1.0*10−14 we know that: + − −log [H3O ] + −log [OH ] = −log Kw = 14.00 or, in other words: pH + pOH = pKw = 14.00
pH and pOH • For any other type solution, the hydrogen or hydroxide ion concentrations will depend on BOTH the dissociation of water and ions contributed by other components of a solution. • For example, if we make a 0.20 Mol/L solution of nitric acid the hydrogen ion concentration would depend on the hydrogen ion from the nitric acid and from the dissociation of water.
HOW DO WE MEASURE PH? (1) For less accurate measurements, one can use Litmus paper: • Red paper turns blue above ~pH = 8 • Blue paper turns red below ~pH = 5 Or use an indicator
HOW DO WE MEASURE PH? (2) For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.
PH: AN EXAMPLE TO CONSIDER (1) • What is the pH of a solution containing 0.30 M HCOOH (pKa = 3.77) and 0.52 M HCOO-?
Solution: Here we have a mixture of weak acid and conjugate base! Let’s set up ICE Table HCOOH (aq) D H+ (aq) + HCOO- (aq)
Initial
Change Equilibrium
0.30
0.00
0.52
-x
+x
+x
0.30 - x
x
0.52 + x
PH: AN EXAMPLE TO CONSIDER (2) • What is the pH of a solution containing 0.30 M HCOOH (pKa = 3.77) and 0.52 M HCOO-? Solution: lets calculate pH then HCOOH (aq) D H+ (aq) + HCOO- (aq)
Equilibrium
0.30 - x
pH pK a
x
HCOO log
0.52 + x
HCOOH
Considering: 0.30 – x 0.30 & 0.52 + x 0.52 0.52 pH 3.77 log 4.01 0.30
AN IMPORTANT NOTE • How do you know when you can use this simplifying assumption ([A]-x [A])? • It can be shown that if the acid concentration [A] divided by the Ka exceeds 100, that is:
[ A ] if
[ A ] 100 or 5% Ka [HA ]
• Then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%. • If the simplifying assumption is not valid, you can solve the equilibrium equation exactly by using the quadratic equation.
SUMMARIZING pH AND pOH As [H3O+] rises, [OH-] falls
OH-
OH
+ H 0
1
2
Acidic
3
OH
+ H 4
5
6
7
8
H+ 9
10
11
Neutral As pH falls, pOH rises
12
13
14
Basic
ACIDS STRENGTH • Strong acids are, by definition, strong electrolytes and exist totally as ions in aqueous solution, whereas weak acids are partially dissociated in water. • You will recall that the strong acids are 7: 3 hydrohalic acids: HCl, HBr, HI 4 Oxo-acids: HNO3, H2SO4, HClO3, & HClO4 • Some weak acids are: Hydrohalic acid: HF With H not bonded to O or hal.: HCN, H2S Oxo-acids: HClO, HNO2, H3PO4 Carboxylic acids: CH3COOH, HCOOH, etc.
BASES STRENGTH
• Strong bases are the soluble hydroxides. Again, these substances dissociate completely in aqueous solution. • Weak bases are treated similarly to weak acids. Some example are: • Strong bases: M2O & MOH M = Group 1A (Li,Na,K,Rb,Cs) MO & M(OH) 2 M = Group 2A (Ca, Sr, Ba) MgO and Mg(OH)2 are only slightly soluble • Weak bases, usually has N atom and lone pair of electrons: NH3 Amines
AN EXAMPLE TO CONSIDER Classify each of the following compounds as a strong acid, weak acid, strong base, or weak base. a) H2SO4 b) (CH3)2CHCOOH c) KOH d) (CH3)2CHNH2 Plan: Pay attention to the text definitions of acids and bases. Look at O for acids as well as the -COOH group; watch for amine groups and cations in bases. Solution: a)Strong acid - H2SO4 b)Weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group. c)Strong base - KOH is a Group 1A hydroxide. d)Weak base - (CH3)2CHNH2 has a lone pair of electrons on the N and is an amine.
DISSOCIATION CONSTANTS (1) • For a generalized acid dissociation:
HA(aq) + H2O(l) D
− A (aq)
+
+ H3O (aq)
the equilibrium expression would be:
[H Ka = 3
+ O]
− [A ]
[HA]
• This equilibrium constant is called the acid-dissociation constant (Ka).
DISSOCIATION CONSTANTS (2) • Acid strength describes the position of the equilibrium or the size of Ka, for the general acid dissociation: + H * X HX(aq) D H (aq) + X (aq) Ka HX • For example: H2SO4 H+ + HSO4- Ka ~ 1.0*103 HSO4H+ + SO4- Ka ~ 1.2*10-2 • In the first case the reaction essentially goes to completion, in the second HSO4- hardly dissociates.
DISSOCIATION CONSTANTS (3) The greater the value of Ka, the stronger the acid.
A Strong Acid
A Weak Acid
Graphical representation of the behavior of acids of different strengths in aqueous solution.
VARIOUS WAYS TO DESCRIBE ACIDS STRENGTH
DISSOCIATION CONSTANTS (4) AN EXAMPLE TO CONSIDER • The pH of a 0.10 M solution of formic acid (HCOOH) at 25°C is 2.38. Calculate Ka for formic acid at this temperature. • We know that: +] [COO−] [H O 3 Ka = [HCOOH]
DISSOCIATION CONSTANTS (5) AN EXAMPLE TO CONSIDER • The pH of a 0.10 M solution of formic acid (HCOOH) at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Solution: • To calculate Ka, we need the equilibrium concentrations of all three species: H3O+, HCOO− and HCOOH. • We can find [H3O+], which is the same as [HCOO−], from the pH.
DISSOCIATION CONSTANTS (6) AN EXAMPLE TO CONSIDER • The pH of a 0.10 M solution of formic acid (HCOOH) at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Solution: pH = −log [H3O+] 2.38 = −log [H3O+] −2.38 = log [H3O+] 10−2.38 = [H3O+] 4.2*10−3 = [H3O+] = [HCOO−]
DISSOCIATION CONSTANTS (7) AN EXAMPLE TO CONSIDER • The pH of a 0.10 M solution of formic acid (HCOOH) at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Solution: Now set up the ICE Table Initially
[HCOOH] 0.10
Change
−4.2*10-3
Equilibrium
0.0958
[H3O+] 0
[HCOO−] 0
+4.2*10-3 +4.2*10−3
4.2*10−3
4.2*10−3
DISSOCIATION CONSTANTS (8) AN EXAMPLE TO CONSIDER • The pH of a 0.10 M solution of formic acid (HCOOH) at 25°C is 2.38. Calculate Ka for formic acid at this temperature. Solution: Now calculate Ka
[4.2*10−3] [4.2*10−3] Ka = [0.0958]
= 1.8*10−4
DISSOCIATION CONSTANTS (9) RELATIONSHIP BETWEEN Ka AND PKa • As we know: pKa = -logKa • Then, a low pKa corresponds to a high value of Ka. Ka at 25°C
pKa
Hydrogen sulfate ion (HSO4-)
1.0x10-2
1.99
Nitrous acid (HNO2)
7.1x10-4
3.15
Acetic acid (CH3COOH)
1.8x10-5
4.75
Hypobromous acid (HBrO)
2.3x10-9
8.64
Phenol (C6H5OH)
1.0x10-10
10.00
Acid Name (Formula)
DISSOCIATION CONSTANTS (10) WEAK BASES • Equilibria involving weak bases are treated similarly that for weak acids. B (aq) + H2O (l) D BH+ (aq) + OH- (aq) • The base-dissociation or baseionization constant is given by:
BH * OH
Kb
B
• Note that no base actually dissociates in solution, but ions are produced when the base reacts with H2O.
SUMMARIZING DISSOCIATION CONSTANTS • Relative strengths of weak acids and bases can be determined from the value of the equilibrium constant. Large equilibrium constant means strong acids and/or bases. Small equilibrium constant means weak acids and/or bases.
General Strategy for solving Acid-Base Problems (1) 1.THINK CHEMISTRY: Focus on the solution components and their reactions. It will almost always be possible to choose one reaction that is the most important. 2.BE SYSTEMATIC: Acid - Base Problems require a step-by-step approach. 3.BE FLEXIBLE: Although all Acid - Base Problems are similar in many ways, important differences do occur. Treat each problem as a separate entity. Do not try to force a given problem to match any you have solved before. Look for both the similarities and the differences.
General Strategy for solving Acid-Base Problems (2) 4. BE PATIENT: The complete solution to a complicated problem cannot be seen immediately in all its detail. Pick the problem apart into its workable steps. 5. BE CONFIDENT: Look within the problem for the solution, and let the problem guide you. Assume that you can think it out. Do not rely on memorizing solutions to problems. In fact, memorizing solutions is usually detrimental, because you tend to try to force a new problem to be the same as one you have seen before. Understand and think; don’t just memorize.
CONCLUSIONS 1.pH can be measure using pH meter, pH paper or acid-base indicator. 2.Since acids and bases are opposites, pH and POH are opposites! 3.“Strong” and “weak” are used in the same sense for bases as for acids. 4.Strong acids and bases completely dissociate in water whereas Weak acids and bases only partially dissociate. THANK YOU!!!
