Chapter 16 Acid–Base Equilibria

Chapter 16 Acid–Base Equilibria
Lecture Presentation Chapter 16 Acid–Base Equilibria John D. Bookstaver St. Charles Community College Cottleville, MO © 2012 Pearson Education, Inc.

Acid base solutions Phet sim
Neutralization – Bozeman #30 pH scale – Phet simulation pH scale – Phet sim II Acid Base Equilibrium – Bozeman #68 pH and Buffers – Bozeman #69 Acid Base reactions in solution – Crash course #8 pH and pOH – crash course #30

16.1 Acid review Arrhenius HCl(aq)  H+ + Cl-
An acid when dissolved in water, increases the concentration of hydrogen ions. We think of acids as H+ generators H+ is sometimes referred to as a “proton” HCl(aq)  H+ + Cl-

16.1 Acid review Arrhenius NaOH(s)  Na+(aq) + OH-(aq)
A base when dissolved in water, increases the concentration of hydroxide ions. NaOH(s)  Na+(aq) + OH-(aq) The presence of excess H+ or OH- ions give acids and bases their characteristic properties Ex: acids taste sour; bases taste bitter

16.2 Bronsted-Lowry acid/base
We now recognize that the H+ ion (really a single proton) must be pulled off by a neighboring molecule with a spare pair of electrons to attach itself to. H3O+ the hydronium ion Water acts as the H+ acceptor when acids are aqueous. This gives rise to a second definition for acids.

Brønsted–Lowry acids and bases
An acid is a proton (H+) donor. A base is a proton acceptor. H+ BL acid Notice HNO2 is also an Arrhenius acid since you now have extra H3O+ ions in the solution H+ Notice NH3 is also an Arrhenius base since we now have extra OH- ions in the solution BL base

Brønsted–Lowry acids and bases
Wherever there is a B-L acid, there is also a B-L base. They come in pairs. Wherever there is a B-L base, there is also a B-L acid. They come in pairs. H+ donor H+ H+ receiver BL acid + BL base Wherever H+ are donated, they must be received H+ BL base + BL acid Wherever H+ is received, it must be donated H+ receiver H+ donor

Ammonia is a Bronsted base because it accepts a proton.
It is an Arrehnius base because it creates OH- (aq)

A Brønsted–Lowry acid…
…must have a removable (acidic) proton.

A Brønsted–Lowry base…
…must have a pair of nonbonding electrons.

If it can be either…it is amphiprotic
HCO3- has one H to lose, but can also gain an H Acid: HCO3 + X-  CO32- + HX Base: HCO3 + HY  H2CO3 + Y- Water is both the acid and base when it self ionizes H2O + H2O  HO- + H3O+ Water is amphiprotic © 2012 Pearson Education, Inc.

Acid base Pairs of acids and bases like this are found on the same side

What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted–Lowry base and abstracts a proton (H+) from the acid. As a result, the hydronium and chloride ions are formed. The Cl- ion is called the conjugate base. acid “conjugate base"

Conjugate Acids and Bases
The term conjugate comes from the Latin word “conjugare,” meaning “to join together.” Reactions between acids and bases always yield their conjugate bases and acids. These pairs are different from before. They are found on opposite sides of the equations

How does an acid become a “base” and a base become an “acid”?
Look at the reverse reaction: ... and becomes a conjugate acid In the reverse reaction the base receives an H+ from H3O+ And the acid becomes a base!

Sample Exercise 16.1 Identifying Conjugate Acids and Bases
(a) What is the conjugate base of HClO4, H2S, PH4+, HCO3? (b) What is the conjugate acid of CN, SO42, H2O, HCO3? Solution Analyze We are asked to give the conjugate base for several acids and the conjugate acid for several bases. Plan The conjugate base of a substance is simply the parent substance minus one proton, and the conjugate acid of a substance is the parent substance plus one proton. Solve HClO4 less one proton H+ is ClO4. The other conjugate bases are HS, PH3, and CO32. 16

(b) What is the conjugate acid of CN, SO42, H2O, HCO3?
(b) CN plus one proton H+ is HCN. The other conjugate acids are HSO4, H3O+, and H2CO3. Notice that the hydrogen carbonate ion (HCO3) is amphiprotic. It can act as either an acid or a base. Practice Exercise Write the formula for the conjugate acid of each of the following: HSO3, F, PO43, CO. H2SO3, HF, HPO42, HCO+

(b) in which the ion acts as a base.
Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions The hydrogen sulfite ion (HSO3) is amphiprotic. Write an equation for the reaction of HSO3 with water (a) in which the ion acts as an acid and (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. Solution Solve The conjugate pairs in this equation are HSO3 (acid) and SO32 (conjugate base), and H2O (base) and H3O+ (conjugate acid). 18

In both cases identify the conjugate acid–base pairs.
The hydrogen sulfite ion (HSO3) is amphiprotic. Write an equation for the reaction of HSO3 with water (b) in which the ion acts as a base. In both cases identify the conjugate acid–base pairs. (b) The conjugate pairs in this equation are H2O (acid) and OH (conjugate base), and HSO3- (base) and H2SO3 (conjugate acid).

O2(aq) + H2O(l)  OH(aq) + OH(aq).
Sample Exercise 16.2 Writing Equations for Proton-Transfer Reactions Continued Practice Exercise When lithium oxide (Li2O) is dissolved in water, the solution turns basic from the reaction of the oxide ion (O2) with water. Write the equation for this reaction and identify the conjugate acid–base pairs. base Acid ….and conjugate base …and conjugate “acid” Answer: O2(aq) + H2O(l)  OH(aq) + OH(aq). The OH is both the conjugate acid of O2 and the conjugate base of H2O. 20

Acid / base strength Oxymoronically, (is that a word?) strong acids produce a lot of H+ since they can’t hold on to their H’s and lose them easily to a base like H2O. (Their “base” is weak?) They’re strong since a little acids gives a lot of H+ Their base (…that’s conjugate base to you) Must be fairly weak, wouldn’t you say? Ex: the Cl- in HCl must itself be a weak base. True?

Weak acids yield little of their H’s to nearby H2O’s
They must be stronger conjugate bases to give up so little of their H. Ex: in a weak acid such as acetic acid HC2H3O2, (think vinegar acid ). The acetate ion itself must be strong enough to hold onto most of the H’s.

Acid and Base Strength Strong acids are completely dissociated in water. Their conjugate bases are extremely weak. Weak acids only dissociate partially in water. Their conjugate bases are also weak bases (but not as weak as above! - they can take the H+ back)

Acid / Base equilibria Strong acid HCl + H2O  HO- + H3O+
Strong acid Completely ionized Keq = extremely large value Negligible base Cl- + H2O  HCl + OH- Negligible no ionization base Keq = extremely small value

Acid / Base equilibria Weak acid HC2H3O2 + H2O  C2H3O2- + H3O+
Weak acid slightly ionized Keq = 1.8 x 10-5 Weak base C2H3O H2O  HC2H3O OH- Weak base slightly ionized Keq = 5.6 x 10-10 Notice that a weak acid has inside itself a weak conjugate base The weak conjugate base will act to ionize water but also to only a small extent

Acid and Base Strength  In any acid–base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. (think: “tug of war” between bases) HCl(aq) + H2O(l)  H3O+(aq) + Cl(aq) Strong acid (weak CB) stronger base Weak acid Weaker base Notice a “strong” side … and a weak side H2O is a much stronger base than Cl, so the equilibrium lies so far to the right that K is not measured (K >> 1). K is “very large”

Acid and Base Strength In any acid–base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base: CH3COOH(aq) + H2O(l) H3O+(aq) + CH3COO(aq) Weak acid weaker base Stronger acid Stronger base Acetate is a stronger base than H2O, so the equilibrium favors the left side (K < 1).

Sample Exercise 16.3 Predicting the Position of a Proton-Transfer Equilibrium
For the following proton-transfer reaction use Figure 16.3 to predict whether the equilibrium lies to the left (Kc < 1) or to the right (Kc > 1): 29

Solution Solve The CO32 ion appears lower in the right-hand column in Figure 16.3 and is therefore a stronger base than SO42. CO32, therefore, will get the proton preferentially to become HCO3, while SO42 will remain mostly unprotonated. The resulting equilibrium lies to the right, favoring products (that is, Kc > 1): Comment Of the two acids HSO4 and HCO3, the stronger one (HSO4) gives up a proton more readily, and the weaker one (HCO3) tends to retain its proton. Thus, the equilibrium favors the direction in which the proton moves from the stronger acid and becomes bonded to the stronger base.

Practice Exercise (a) left, (b) right
For each reaction, use Figure 16.3 to predict whether the equilibrium lies to the left or to the right: (a) (b) Answers: (a) left, (b) right 31

16.3 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. This is referred to as autoionization.

Ion Product Constant 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 25C, Kw = 1.0  1014 (Since the reactant is a liquid it is left out of the Keq expression) You can tell that very few ions are formed since this value is soooo small

In an acid solution [H+] is greater than 1.0  107 M;
Sample Exercise 16.4 Calculating [H+] for Pure Water Calculate the values of [H+] and [OH] in a neutral solution at 25 C . Solution Analyze We are asked to determine the concentrations of H+ and OH ions in a neutral solution at 25 C . Plan We will use Equation and the fact that, by definition, [H+] = [OH] in a neutral solution. Solve We will represent the concentration of H+ and OH in neutral solution with x. This gives In an acid solution [H+] is greater than 1.0  107 M; in a basic solution [H+] is less than 1.0  107 M. 34

Practice Exercise Indicate whether solutions with each of the following ion concentrations are neutral, acidic, or basic: [H+] = 4  109 M; [OH] = 1  107 M; [OH] = 7  1013 M. Answers: (a) basic, (b) neutral, (c) acidic

Sample Exercise 16.5 Calculating [H+] from [OH]
Calculate the concentration of H+ (aq) in a solution in which [OH] is M, a solution in which [OH] is 1.8  109 M. Note: In this problem and all that follow, we assume, unless stated otherwise, that the temperature is 25 C . Solution Plan We can use the equilibrium-constant expression for the autoionization of water and the value of Kw to solve for each unknown concentration. Solve (a) Using Equation 16.16, we have This solution is basic because [OH] > [H+] 36

Calculate the concentration of H+ (aq) in
(b) a solution in which [OH] is 1.8  109 M. . (b) In this instance This solution is acidic because [H+] > [OH]

Practice Exercise Answers: (a) 5  109 M, (b) 1.0  107 M,
Sample Exercise 16.5 Calculating [H+] from [OH] Continued Practice Exercise Calculate the concentration of OH(aq) in a solution in which (a) [H+] = 2  106 M; (b) [H+] = [OH]; (c) [H+] = 100  [OH]. Answers: (a) 5  109 M, (b) 1.0  107 M, (c) 1.0  108 M 38

pH Recall, In pure water, Kw = [H3O+] [OH] = 1.0  1014
It should be pointed out that this Kw is for water at 250C. Kw increases slightly at higher temperatures. Recall, In pure water, Kw = [H3O+] [OH] = 1.0  1014 Since in pure water, [H3O+] = [OH], [H3O+] = 1.0  1014 = 1.0  107 pH = log [H3O+] = log (1.0  107) = 7.00 [10 -7][10 -7 ] = 1014 Literally “what is the negative of the base 10 exponent in the number 10-7? Duh. Try it on your calculator

Also, Remember the equilibrium?:
and that Kw = [H3O+] [OH] = 1.0  1014 As [H3O+] increases, [OH] must decrease Kw = [H3O+][OH] = 1.0  1014 This can be confusing: since the exponent is negative, the value gets larger as the exponent gets smaller! Ex: [10 -5 ] [10 -9] = 1.0  1014 pH = log [H3O+] = log (105) = 5.00 An acid has a higher [H3O+] than pure water, so its pH is lower <7. 10-5 > 10-7

For bases pH = log [H3O+] = log (109) = 9.00
As [OH] increases, [H3O+] decreases, Kw = [H3O+][OH] = 1.0  1014 Ex: [10 -9 ][10 -5] = 1.0  1014 pH = log [H3O+] = log (109) = 9.00 A base has a lower [H3O+] than pure water, so its pH is higher > 7. Bases have small [H+], so they’re [OH-] must be larger

[H3O+][OH] [H3O+][OH] [H3O+][OH]
Figure 16.4 Relative concentrations of H+ and OH– in aqueous solutions at 25 °C. 43

These are the pH values for several common substances Fig. 16.5.
Human blood is slightly basic to help absorb the lactic acid that streams into your veins during a hard workout! Stomach Acid would be considered a hazardous substance outside the body! These are the pH values for several common substances Fig

Solution Solve Sample Exercise 16.6 Calculating pH from [H+]
Calculate the pH values for the two solutions of Sample Exercise 16.5. a solution in which [OH] is M, a solution in which [OH] is 1.8  109 M. a) Solution Plan We can calculate pH using its defining equation, Equation Solve (a) In the first instance we found [H+] to be 1.0  1012 M, so that pH = log(1.0  1012) = (12.00) = 12.00 Because 1.0  1012 has two significant figures, the pH has two decimal places,

(b) a solution in which [OH] is 1.8  109 M. .
(b) For the second solution, [H+] = 5.6  106 M . Before performing the calculation, it is helpful to estimate the pH. To do so, we note that [H+] lies between 1  106 and 1  105. Thus, we expect the pH to lie between 6.0 and 5.0. We use Equation to calculate the pH: pH = log(5.6  106) = 5.25 Check After calculating a pH, it is useful to compare it to your estimate. In this case the pH, as we predicted, falls between 6 and 5. Had the calculated pH and the estimate not agreed, we should have reconsidered our calculation or estimate or both.

Practice Exercise In a sample of lemon juice, [H+] = 3.8  104 M. What is the pH? A commonly available window-cleaning solution has [OH] = 1.9  106 M. What is the pH? Answers: (a) 3.42, (b) [H+] = 5.3  109 M, so pH = 8.28

Other “p” Scales The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions). Some similar examples are pOH: log [OH] pKw: log Kw © 2012 Pearson Education, Inc.

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 Again…at 250C

Various pH indicator solutions and their colors.
Figure 16.7 pH ranges for common acid–base indicators. These are the pH ranges over which the color changes. 51

Universal indicator

Regents reference table for indicators:

Sample Exercise 16.7 Calculating [H+] from pOH
A sample of freshly pressed apple juice has a pOH of Calculate [H+]: 55

Solution Plan We will first use Equation 16.20, pH + pOH = 14.00, to calculate pH from pOH. Then we will use Equation to determine the concentration of H+. Solve From Equation 16.20, we have pH =  pOH pH = – = 3.76 Next we use Equation 16.17: pH = log [H+] = 3.76 Thus, log [H+] = 3.76 To find [H+], we need to determine the antilog of 3.76. Scientific calculators have an antilogarithm function (sometimes labeled INV log or 10x) that allows us to perform the calculation: [H+] = antilog (3.76) = 103.76 = 1.7 × 104 M Recall, we used ln x and ex functions when dealing with reaction orders?

Sample Exercise 16.7 Calculating [H+] from pOH
Continued Comment The number of significant figures in [H+] is two because the number of decimal places in the pH is two. Check Because the pH is between 3.0 and 4.0, we know that [H+] will be between 1.0  103 and 1.0  104M. Our calculated [H+] falls within this estimated range. Practice Exercise A solution formed by dissolving an antacid tablet has a pH of Calculate [OH]. Answer: [OH] = 1.5  105 57

6.5 Strong Acids You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4. These are, by definition, strong electrolytes and exist totally as ions in aqueous solution. For the monoprotic strong acids, [H3O+] = [acid]. Ex: 0.10 M HCl = 0.10 M H+

Sample Exercise 16.8 Calculating the pH of a Strong Acid
What is the pH of a M solution of HClO4? Solution Analyze and Plan Because HClO4 is a strong acid, it is completely ionized, giving HClO4(aq)  H+ (aq) + ClO4- (aq) [HClO4] = [H+] = [ClO4] = M. Solve pH = log(0.040) = 1.40 Practice Exercise An aqueous solution of HNO3 has a pH of 2.34.What is the concentration of the acid? Answer: M 59

Strong Bases Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+ - Remember “CBS”). Again, these substances dissociate completely in aqueous solution. © 2012 Pearson Education, Inc.

Sample Exercise 16.9 Calculating the pH of a Strong Base
What is the pH of a M solution of NaOH, a M solution of Ca(OH)2? 61

Solution Analyze We are asked to calculate the pH of two solutions of strong bases. Plan We can calculate each pH by either of two equivalent methods. First, we could use Equation to calculate [H+]and then use Equation to calculate the pH. Alternatively, we could use [OH] to calculate pOH and then use Equation to calculate the pH. Solve (a) NaOH dissociates in water to give one OH ion per formula unit. Therefore, the OH concentration for the solution in (a) equals the stated concentration of NaOH, namely M. Method 1: Method 2: pOH = log(0.028) = pH = – pOH = 12.45

pH pOH ? [OH-] [H+]

Practice Exercise b) What is the pH of a 0.0011 M solution of Ca(OH)2?
(b) Ca(OH)2 is a strong base that dissociates in water to give two OH ions per formula unit. Thus, the concentration of OH(aq) for the solution in part (b) is 2  ( M) = M. Method 1: Method 2: pOH = log(0.0022) = pH =  pOH = 11.34 Practice Exercise What is the concentration of a solution of (a) KOH for which the pH is 11.89, (b) Ca(OH)2 for which the pH is 11.68? Answer: (a) 7.8  103 M, (b) 2.4  103 M 64

16.6 Dissociation Constants: weak acids
For a generalized acid dissociation, the equilibrium expression would be This equilibrium constant is called the acid-dissociation constant, Ka. HA(aq) + H2O(l) A(aq) + H3O+(aq) [H3O+] [A] [HA] Kc = © 2012 Pearson Education, Inc.

Dissociation Constants
The greater the value of Ka, the stronger is the acid.

Calculating Ka from the pH
The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. We know that HCOOH(aq)   H+(aq) + HCOO-(aq) [H3O+] [HCOO] [HCOOH] Ka =

The pH of a 0.10 M solution of formic acid, HCOOH, at 25C is Calculate Ka for formic acid at this temperature. To calculate Ka, we need the equilibrium concentrations of all three things. We can find [H3O+], which is the same as [HCOO], from the pH. [H3O+] [HCOO] [HCOOH] Ka = © 2012 Pearson Education, Inc.

Calculating Ka from pH HCOOH(aq)   H+(aq) + HCOO-(aq)
Now we can set up an ICE table… [HCOOH], M [H3O+], M [HCOO], M Initially 0.10 Change At equilibrium HCOOH(aq)   H+(aq) + HCOO-(aq)

Calculating Ka from pH Now we can set up a table… [HCOOH], M [H3O+], M [HCOO], M Initially 0.10 Change 4.2  103 +4.2  103 At equilibrium HCOOH(aq)  H+(aq) + HCOO-(aq) © 2012 Pearson Education, Inc.

Calculating Ka from pH HCOOH(aq)  H+(aq) + HCOO-(aq)
Now we can set up a table… [HCOOH], M [H3O+], M [HCOO], M Initially 0.10 Change 4.2  103 +4.2  103 At equilibrium 0.10  4.2  103 = = 0.10 4.2  103 Notice: Doesn’t change initial concentration “significantly” HCOOH(aq)  H+(aq) + HCOO-(aq)

Practice Exercise Answer: 1.5  105
Niacin, one of the B vitamins, has the molecular structure A M solution of niacin has a pH of 3.26. What is the acid-dissociation constant for niacin? Answer: 1.5  105

Calculating Percent Ionization
[H3O+]eq [HA]initial Percent ionization =  100 In this example, [H3O+]eq = 4.2  103 M [HCOOH]initial = 0.10 M 4.2  103 0.10 Percent ionization =  100 = 4.2% © 2012 Pearson Education, Inc.

Sample Exercise 16.11 Calculating Percent Ionization
As calculated in Sample Exercise 16.10, a 0.10 M solution of formic acid (HCOOH) contains 4.2  103 M H+(aq). Calculate the percentage of the acid that is ionized. 76

Solution Practice Exercise
Analyze We are given the molar concentration of an aqueous solution of weak acid and the equilibrium concentration of H+ (aq) and asked to determine the percent ionization of the acid. Plan The percent ionization is given by Equation Solve Practice Exercise A M solution of niacin has a pH of Calculate the percent ionization of the niacin. Answers: 2.7%

Calculating pH from Ka Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2(aq) Ka for acetic acid at 25C is 1.8  105. © 2012 Pearson Education, Inc.

Calculating pH from Ka We next set up a table…
[C2H3O2], M [H3O+], M [C2H3O2], M Initially 0.30 Change x +x At equilibrium 0.30  x  0.30 x We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored. (“Since Ka is small there is not significant ionization”) Sneaky, but allows us to avoid the quadratic equation. You must mention it on the AP exam to receive credit though.

Solution Sample Exercise 16.12 Using Ka to Calculate pH
Calculate the pH of a 0.20 M solution of HCN. (Refer to Table 16.2 or Appendix D for the value of Ka.) Solution Plan We proceed as in the example just worked in the text, writing the chemical equation and constructing a table of initial and equilibrium concentrations in which the equilibrium concentration of H+ is our unknown. 85

Sample Exercise 16.12 Using Ka to Calculate pH
Solve Writing both the chemical equation for the ionization reaction that forms H+ (aq) and the Equilibrium-constant (Ka) expression for the reaction: Next, we tabulate the concentrations of the species involved in the equilibrium reaction, letting x = [H+] at equilibrium:

We now calculate the pH of the solution:
Substituting the equilibrium concentrations into The equilibrium-constant expression yields We next make the simplifying approximation that x, the amount of acid that dissociates, is small compared with the initial concentration of acid, Thus, Solving for x, we have A concentration of 9.9  106 M H+ is much smaller than 5% of 0.20, the initial HCN concentration. Our simplifying approximation is therefore appropriate. We now calculate the pH of the solution: 87

Practice Exercise The Ka for niacin (Practice Exercise 16.10) is 1.5  105. What is the pH of a M solution of niacin? Answers: 3.41 88

Sample Exercise 16.13 Using Ka to Calculate Percent Ionization
Calculate the percentage of HF molecules ionized in a 0.10 M HF solution, a M HF solution.

A quick check: ________________ x 100 = 8.2 % 0.10
Solution Plan We approach this problem as we have previous equilibrium problems. We begin by writing the chemical equation for the equilibrium and tabulating the known and unknown concentrations of all species. We then substitute the equilibrium concentrations into the equilibrium-constant expression and solve for the unknown concentration, that of H+. Solve (a) The equilibrium reaction and equilibrium concentrations are as follows: The equilibrium-constant expression is When we try solving this equation using the approximation (that is, by neglecting the concentration of acid that ionizes), we obtain A quick check: ________________ x 100 = 8.2 % 0.10 90

Substituting these values in the standard quadratic formula gives
Because this value is greater than 5% of 0.10 M, however, we should work the problem without the approximation. Rearranging our equation and writing it in standard quadratic form, we have Substituting these values in the standard quadratic formula gives Of the two solutions, only the one that gives a positive value for x is chemically reasonable. Thus,

From our result, we can calculate the percent of Molecules ionized:
92

Sample Exercise 16.13 Using Ka to Calculate Percent Ionization
Continued (b) Proceeding similarly for the M solution, we have Solving the resultant quadratic expression, we obtain The percentage of molecules ionized is

Comment Notice that if we do not use the quadratic formula, we calculate 8.2% ionization for (a) and 26% ionization for (b). Notice also that in diluting the solution by a factor of 10, the percentage of molecules ionized increases by a factor of 3. This result is in accord with what we see in Figure It is also what we expect from Le Châtelier’s principle. (Section 15.7) There are more “particles” or reaction components on the right side of the equation than on the left. Dilution causes the reaction to shift in the direction of the larger number of particles (toward the product side) because this counters the effect of the decreasing concentration of particles. Ironically, as a weak acid is diluted (it becomes less concentrated), its percent of ionized molecules increases. However, in spite of this the H+ ion concentration still decreases. Notice that the fraction of molecules ionized is fairly low in concentrated solutions.(Water molecules must be present to pull the H+ ions from the acid molecules) 94

Polyprotic Acids Polyprotic acids have more than one acidic proton.
If the difference between the Ka for the first dissociation and subsequent Ka values is 103 or more, the pH generally depends only on the first dissociation.

What is the pH of a 0.0037 M solution of H2CO3?
Sample Exercise Calculating the pH of a Polyprotic Acid Solution The solubility of CO2 in water at 25 C and 0.1 atm is M. The common practice is to assume that all the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced in the reaction What is the pH of a M solution of H2CO3? 96

Solution Plan H2CO3 is a diprotic acid; the two acid-dissociation constants, Ka1 and Ka2 (Table 16.3), differ by more than a factor of 103. Consequently, the pH can be determined by considering only Ka1, thereby treating the acid as if it were a monoprotic acid.

Solve Proceeding as in Sample Exercises 16. 12 and 16
Solve Proceeding as in Sample Exercises and 16.13, we can write the equilibrium reaction and equilibrium concentrations as The equilibrium-constant expression is Alternatively, because Ka1 is small, we can make the simplifying approximation that x is small, so that

Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid Solution
Continued Thus, Solving for x, we have Because we get the same value (to 2 significant figures) our simplifying assumption was justified. ( – 10-5 = ) The pH is therefore pH = log[H+] = log(4.0  105) = = 4.40 99

Sample Exercise 16.14 Calculating the pH of a Polyprotic Acid Solution
Continued Comment If we were asked for [CO32], we would need to use Ka2. Let’s illustrate that calculation. Using our calculated values of [HCO3] and [H+] and setting [CO32] = y, we have Assuming that y is small relative to 4.0  105, we have 100

This value is indeed very small compared with 4
This value is indeed very small compared with 4.0  105, showing that our assumption was justified. It also shows that the ionization of HCO3 is negligible relative to that of H2CO3, as far as production of H+ is concerned. However, it is the only source of CO32, which has a very low concentration in the solution. Our calculations thus tell us that in a solution of carbon dioxide in water, most of the CO2 is in the form of CO2 or H2CO3, only a small fraction ionizes to form H+ and HCO3, and an even smaller fraction ionizes to give CO32. Notice also that [CO32] is numerically equal to Ka2. We can check this mathematically by combining the H+ ions produced in each step: Combined [H+] = y + x [5.6 x M x 10-5 M] [ x 10-5] pH = -log [ x 10-5] = = 4.40

Practice Exercise (a) pH = 1.80, (b) [C2O42] = 6.4  105 M
Sample Exercise Calculating the pH of a Polyprotic Acid Solution Continued Practice Exercise (a) Calculate the pH of a M solution of oxalic acid (H2C2O4). (See Table 16.3 for Ka1 and Ka2.) (b) Calculate the concentration of oxalate ion, [C2O42], in this solution. Answers: (a) pH = 1.80, (b) [C2O42] = 6.4  105 M 102

Weak Bases Weak bases can be of two types:
Weak molecular bases – don’t ionize completely insoluble hydroxides – don’t dissolve well. NH3 + H2O  NH4+ + OH- Mg(OH)2(s)  Mg2+(aq) + 2 OH-(aq) In this section we’ll focus on molecular bases like NH3. Insoluble bases will be examined with solubility in chapter 17.

Weak Bases Kb can be used to find [OH] and, through it, pH.

Solution Sample Exercise 16.15 Using Kb to Calculate OH
Calculate the concentration of OH in a 0.15 M solution of NH3 The Kb for ammonia is 1.8 x 10-5 Solution Plan We will use essentially the same procedure here as used in solving problems involving the ionization of weak acids, that is, write the chemical equation and tabulate initial and equilibrium concentrations.

Solution Solve The ionization reaction and equilibrium-constant expression are The equilibrium concentrations are 113

Inserting these quantities into the equilibrium-constant
expression gives Because Kb is small, the amount of NH3 that reacts with water is much smaller than the NH3 concentration, and so we can neglect x relative to 0.15 M. Then we have

Comment You may be asked to find the pH of a solution of a weak base.
Check The value obtained for x is only about 1% of the NH3 concentration, 0.15 M. Therefore, neglecting x relative to 0.15 was justified. Comment You may be asked to find the pH of a solution of a weak base. Once you have found [OH], you can proceed as in Sample Exercise 16.9, where we calculated the pH of a strong base. In the present sample exercise, we have seen that the 0.15 M solution of NH3 contains [OH] = 1.6  103 M. Thus, pOH = log(1.6  103) = 2.80, and pH = – 2.80 = The pH of the solution is above 7 because we are dealing with a solution of a base. 115

Answers: Practice Exercise
Which of the following compounds should produce the highest pH as a 0.05 M solution: pyridine, methylamine, or nitrous acid? Answers: methylamine (because it has the larger Kb value of the two amine bases in the list)

Sample Exercise 16.16 Using pH to Determine the Concentration of a Salt
A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution, has a pH of Using the information in Equation 16.37, calculate the number of moles of NaClO added to the water. (Notice in this case that the salt anion ClO- is acting as a B-L base …Not an Arrhenius base since OH- is not the only negative ion in the solution)

Solution Analyze NaClO is an ionic compound consisting of Na+ and ClO ions. As such, it is a strong electrolyte that completely dissociates in solution into Na+, a spectator ion, and ClO ion, a weak base with Kb = 3.3  107 (Equation 16.37). Given this information we must calculate the number of moles of NaClO needed to raise the basicity of 2.00-L of water to

We can calculate [ClO] using the expression for Kb.
Plan From the pH, we can determine the equilibrium concentration of OH-. We can then construct a table of initial and equilibrium concentrations in which the initial concentration of ClO is our unknown. We can calculate [ClO] using the expression for Kb. Solve We can calculate [OH] by using either Equation 16.16 or Equation 16.20; we will use the latter method here: pOH =  pH =  = 3.50 [OH] = 103.50 = 3.2  104 M 119

(This concentration is high enough that we can assume that Equation 16
(This concentration is high enough that we can assume that Equation is the only source of OH; that is, we can neglect any OH produced by the auto-ionization of H2O.) We now assume a value of x for the initial concentration of ClO and solve the equilibrium problem in the usual way. We now use the expression for the base-dissociation constant to solve for x: ClO-

0.31 M ClO- = 0.31 M NaClO Practice Exercise 0.12 M
We say that the solution is 0.31 M in NaClO even though some of the ClO ions have reacted with water. Molarity = moles solute / liters of solution 0.31M = x moles / 2.00 liters x = 0.62 moles NaClO Because the solution is 0.31 M in NaClO and the total volume of solution is 2.00 L, 0.62 mol of NaClO is the amount of the salt that was added to the water. Practice Exercise What is the molarity of an aqueous NH3 solution that has a pH of 11.17? Answers: 0.12 M 121

Sample Exercise 16.17 Calculating Ka or Kb for a Conjugate Acid–Base Pair
Calculate (a) Kb for the fluoride ion, (b) Ka for the ammonium ion.

Solution Kb = Kw/Ka = 1.0 x 10-14 / 6.8 x 10-4 = 1.5 x 10-11
Analyze We are asked to determine dissociation constants for F, the conjugate base of HF, and NH4+, the conjugate acid of NH3. Plan We can use the tabulated K values for HF and NH3 and the relationship between Ka and Kb to calculate the ionization constants for their conjugates, F and NH4+. Solve (a) For the weak acid HF, Table 16.2 and Appendix D give Ka = 6.8  104. We can use Equation to calculate Kb for the conjugate base, F: Kb = Kw/Ka = 1.0 x / 6.8 x 10-4 = 1.5 x 10-11 124

(b) For NH3, Table 16. 4 and in Appendix D give Kb = 1
(b) For NH3, Table 16.4 and in Appendix D give Kb = 1.8  105, and this value in Equation gives us Ka for the conjugate acid, NH4+: Ka = Kw/Kb = 1.0 x 10-14 1.8 x 10-5 = 5.6 x 10-10 Other common weak bases include the organics; Pyridines, and amines Plus the obvious conjugate bases of weak acids such as H2S, H2CO3 and HClO 125

Practice Exercise PO43 (Kb = 2.4  102), (b) Kb = 7.9  1010
Check The respective K values for F and NH4+ are listed in Table 16.5, where we see that the values calculated here agree with those in Table 16.5. Practice Exercise Which of these anions has the largest base-dissociation constant: NO2, PO43, or N3 (Ka for HNO2 = 4.5x10-4; HN3 = 1.9x10-5; H3PO4 = k1=7.5x10-3; k2=4.2x10-8; k3=4.2x10-13 )? PO43 (Kb = 2.4  102), (b) The base quinoline has the structure Its conjugate acid is listed in handbooks as having a pKa of 4.90.What is the base-dissociation constant for quinoline? (b) Kb = 7.9  1010 126

16.9 Reactions of Anions with Water
Anions are bases. As such, they can react with water in a hydrolysis reaction to form OH and the conjugate acid: X(aq) + H2O(l) HX(aq) + OH(aq) © 2012 Pearson Education, Inc.

Reactions of Cations with Water
Cations with acidic protons (like NH4+) will lower the pH of a solution. Most metal cations that are hydrated in solution also lower the pH of the solution. © 2012 Pearson Education, Inc.

Attraction between nonbonding electrons on oxygen and the metal causes a shift of the electron density in water. This makes the O–H bond more polar and the water more acidic. © 2012 Pearson Education, Inc.

Effect of Cations and Anions
An anion that is the conjugate base of a strong acid will not affect the pH.. Ex: HCl + H2O  Cl- + H3O + Strong acid Cl H2O  No reaction Negligible Base Salt dissolved in water (NaCl  Na+ + Cl-) (No hydrolysis) neutral

2. An anion that is the conjugate base of a weak acid will increase the pH. Ex: H2CO3 + H2O  HCO3- + H3O + Strong acid HCO3- + H2O  H2CO3 + OH- Weak forms hydroxide Base Salt dissolved in water (NaHCO3  Na+ + HCO3-) (hydrolysis) basic © 2012 Pearson Education, Inc.

3. A cation that is the conjugate acid of a weak base will decrease the pH. Ex: NH3 + H2O  NH4+ + OH- Weak base NH4+ + H2O  NH3 + H3O + Weak forms hydronium acid Salt dissolved in water (NH4Cl  NH4+ + Cl-) (hydrolysis) acidic

Cations of the strong Arrhenius bases will not affect the pH. Ex: NaOH  Na+ + OH- Strong base Na H2O  No reaction Negligible acid Salt dissolved in water (NaCl  Na+ + Cl-) © 2012 Pearson Education, Inc.

5. Other metal ions will cause a decrease in pH. Ex: Al(OH)3  Al OH- Weak base Al H2O  Al(H2O)5OH2+ + H3O+ weak acid Salt dissolved in water (AlCl3  Al3+ + 3Cl-) Hydrated ion acidic

6. When a solution contains both a weak acid and a weak base, the affect on pH depends on the Ka and Kb values. Higher K predominates (NH4)2CO3  2NH4+ + CO32- NH4+ + H2O  NH3 + H3O+ Ka = 5.6 x CO32- + H2O  HCO3- + OH- Kb = 2.1 x 10-4 © 2012 Pearson Education, Inc.

(b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3.
Sample Exercise Determining Whether Salt Solutions are Acidic, Basic, or Neutral Determine whether aqueous solutions of each of these salts are acidic, basic, or neutral: (a) Ba(CH3COO)2, (b) NH4Cl, (c) CH3NH3Br, (d) KNO3, (e) Al(ClO4)3. 137

Solution Analyze We are given the chemical formulas of five ionic compounds (salts) and asked whether their aqueous solutions will be acidic, basic, or neutral. Plan We can determine whether a solution of a salt is acidic, basic, or neutral by identifying the ions in solution and by assessing how each ion will affect the pH. Solve

: (a) Ba(CH3COO)2: This solution contains barium ions and acetate ions. The cation is an ion of a heavy alkaline earth metal and will therefore not affect the pH. The anion, CH3COO, is the conjugate base of the weak acid CH3COOH and will hydrolyze to produce OH ions, thereby making the solution basic (combination 2).

(b) NH4Cl In this solution, NH4+ is the conjugate acid of a weak base (NH3) and is therefore acidic. Cl is the conjugate base of a strong acid (HCl) and therefore has no influence on the pH of the solution. Because the solution contains an ion that is acidic (NH4+) and one that has no influence on pH (Cl), the solution of NH4Cl will be acidic (combination 3).

(c) CH3NH3Br . Here CH3NH3+ is the conjugate acid of a weak base (CH3NH2, an amine) and is therefore acidic, and Br is the conjugate base of a strong acid (HBr) and therefore pH neutral. Because the solution contains one ion that is acidic and one that has no influence on pH, the solution of CH3NH3Br will be acidic (combination 3).

(d) KNO3 (d) This solution contains the K+ ion, which is a cation of group 1A, and the NO3 ion, which is the conjugate base of the strong acid HNO3. Neither of the ions will react with water to any appreciable extent, making the solution neutral (combination 1).

(e) Al(ClO4)3. This solution contains Al3+ and ClO4 ions
(e) Al(ClO4)3. This solution contains Al3+ and ClO4 ions. Cations, such as Al3+, that have a charge of 3 or higher are acidic. The ClO4 ion is the conjugate base of a strong acid (HClO4) and therefore does not affect pH. Thus, the solution of Al(ClO4)3 will be acidic (combination 3).

Practice Exercise Indicate which salt in each of the following pairs forms the more acidic (or less basic) M solution: NaNO3 or Fe(NO3)3, (b) KBr or KBrO, (c) CH3NH3Cl or BaCl2, (d) NH4NO2 or NH4NO3. Answers: (a) Fe(NO3)3, (b) KBr (c) CH3NH3Cl, , (d) NH4NO3 144

Sample Exercise 16.19 Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic
Predict whether the salt Na2HPO4 forms an acidic solution or a basic solution when dissolved in water. Solution Analyze We are asked to predict whether a solution of Na2HPO4 is acidic or basic. This substance is an ionic compound composed of Na+ and HPO42 ions. Plan We need to evaluate each ion, predicting whether it is acidic or basic. Because Na+ is a cation of group 1A, it has no influence on pH. Thus, our analysis of whether the solution is acidic or basic must focus on the behavior of the HPO42 ion. We need to consider the fact that HPO42 can act as either an acid or a base: Of these two reactions, the one with the larger equilibrium constant determines whether the solution is acidic or basic. 145

Solve The value of Ka for Equation is 4.2  1013 (Table 16.3). For Equation 16.46, we must calculate Kb for the base HPO42 from the value of Ka for its conjugate acid, H2PO4, and the relationship Ka  Kb = Kw (Equation 16.40).Using the value Ka(H2PO4) = 6.2  108 from Table 16.3, we have This Kb value is more than 105 times larger than Ka for HPO42; thus, the reaction in Equation predominates over that in Equation 16.45, and the solution is basic. 146

Practice Exercise acidic
Sample Exercise Predicting Whether the Solution of an Amphiprotic Anion Is Acidic or Basic Continued Practice Exercise Predict whether the dipotassium salt of citric acid (K2HC6H5O7) forms an acidic or basic solution in water see Table 16.3 for data). Answers: acidic 147

16.10 Factors Affecting Acid Strength
The more polar the H–X bond and/or the weaker the H–X bond, the more acidic the compound. So acidity increases from left to right across a row and from top to bottom down a group.

Factors Affecting Acid Strength
In oxyacids, in which an –OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid. © 2012 Pearson Education, Inc.

Lewis Acids Lewis bases are defined as electron-pair donors. Anything that could be a Brønsted–Lowry base is a Lewis base. Lewis bases can interact with things other than protons, however. © 2012 Pearson Education, Inc.

Chapter 16 Acid–Base Equilibria

Similar presentations

Presentation on theme: "Chapter 16 Acid–Base Equilibria"— Presentation transcript:

Similar presentations

About project

Feedback

Log in

Auth with social network:

Chapter 16 Acid–Base Equilibria

Similar presentations

Presentation on theme: "Chapter 16 Acid–Base Equilibria"— Presentation transcript:

Similar presentations

About project

Feedback