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Acids and Bases AP Chemistry Seneca Valley Chapter 14 1 1 1 1.

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Presentation on theme: "Acids and Bases AP Chemistry Seneca Valley Chapter 14 1 1 1 1."— Presentation transcript:

1 Acids and Bases AP Chemistry Seneca Valley Chapter 14 1 1 1 1

2 What do you know about acids & bases?

3 Some Definitions Arrhenius
An acid is a substance that, when dissolved in water, increases the concentration of hydrogen ions. A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions. Problem with this definition is that it limits us to aqueous solns.

4 Some Definitions Brønsted-Lowry
An acid is a proton (H+ or H3O+) donor. A base is a proton acceptor. It does not need to contain OH-. HCl + H2O ® Cl- + H3O+ acid base

5 A Brønsted-Lowry acid…
…must have a removable (acidic) proton. A Brønsted-Lowry base… …must have a pair of nonbonding electrons.

6 If it can be either… …it is amphoteric. H2O HCO3- HSO4-

7 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 conjugate base of the acid and a hydronium ion are formed.

8 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.

9 Acid and Base Strength Strong acids are completely dissociated in water. Their conjugate bases are quite weak. Weak acids only dissociate partially in water. Their conjugate bases are weak bases. OH- is the strongest base that can exist in equilibrium in aqueous solution.

10 Acid and Base Strength Substances with negligible acidity do not dissociate in water. Their conjugate bases are exceedingly strong.

11 Strong Acids and Bases Conjugate bases
Using the idea of conjugate bases previously covered arrange the following according to their strength as bases. H2O F- Cl- NO CN-

12 Acid and Base Strength In any acid-base reaction, the equilibrium will favor the reaction that moves the proton to the stronger base. HCl (aq) + H2O (l)  H3O+ (aq) + Cl- (aq) H2O is a much stronger base than Cl-, so the equilibrium lies so far to the right that Ka is not measured (Ka>>1).

13 Acid and Base Strength Again, 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) Acetate is a stronger base than H2O, so the equilibrium favors the left side (Ka<1).

14 Relative Strengths of Acids
Consider two separate aqueous solutions: one of a weak acid HA and one of HCl. Assume 10 molecules of each acid. Draw a picture of what each solution looks like at equilibrium. What are the major species in each beaker? From your pictures, calculate the Ka values of each acid. Order the following from strongest to weakest base and explain: H2O, A-(aq), Cl-(aq).

15 Relative Strengths of Acids
Draw molecular-level pictures of the following: concentrated weak acid dilute weak acid concentrated weak base dilute weak base concentrated strong acid dilute strong acid concentrated strong base dilute strong base

16 Weak Acids

17 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. H2O (l) + H2O (l) H3O+ (aq) + OH- (aq)

18 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

19 The pH Scale In most solutions [H+(aq)] is quite small.
We define pH = -log[H+] = -log[H3O+]. Most pH and pOH values fall between 0 and 14. There are no theoretical limits on the values of pH or pOH. (e.g. pH of 2.0 M HCl is )

20 pH In pure water, Kw = [H3O+] [OH-] = 1.0  10-14
Since in pure water [H3O+] = [OH-], [H3O+] =  = 1.0  10-7

21 pH Therefore, in pure water, pH = -log (1.0  10-7) = 7.00
An acid has a higher [H+] than pure water, so its pH is <7. A base has a lower [H+] than pure water, so its pH is >7.

22 pH These are the pH values for several common substances.

23 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

24 -log [H3O+] + -log [OH-] = -log Kw = 14.00
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

25 How Do We Measure pH?

26 How Do We Measure pH? For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

27 The pH Scale The number of decimal places in the log (pH) is equal to the number of significant figures in the original number. For example: [H+] = 1.0 x 10-9  2 SF pH = 9.00  2 decimal places

28 The pH Scale Solving for [H+], [OH-] and pH 1.0x10-5 M OH-
10.0 M H+ See sample problems 1-3.

29 The pH Scale Measuring pH Calculate the pH of 0.10 M HNO3.
Step 1: List the materials used to prepare solution & the major species in the solution. Step 2: Indicate the major & minor sources of H+. (Write the reactions that produce H+.) Step 3: Indicate the approximations to be made. Step 4: Calculate the pH. See sample problem 4.

30 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].

31 Calculating the pH for solutions containing weak acids
This situation is more complex than for a strong acid. Since a weak acid does not completely dissociate, an equilibrium calculation must be performed to find [H+]. A weak acid can be recognized by its small Ka value.

32 Solving Acid-Base Problems
List the major species. Strong acids are written as completely dissociated. Weak acids are written as HA. Write the balance equations for major species that produce H+ and determine the dominant source of H+. Write the equilibrium expression for the dominant acid. 4. List the initial concentrations of all species involved in the dominant equilibrium.

33 Solving Acid-Base Problems
Define x, the change required to reach equilibrium. Write the equilibrium concentrations in terms of the initial concentrations and x. Substitute the equilibrium concentrations in the expression for Ka. Simplify the expression by neglecting x where possible. That is, assume that [HA] = [HA]0 – x ≈ [HA]0. Then solve for x. Check the validity of the assumption made in step 8. (Use 5% rule). Calculate [H+] and pH.

34 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.

35 Calculating pH from Ka [H3O+] [C2H3O2-] Ka = [HC2H3O2]
The equilibrium constant expression is [H3O+] [C2H3O2-] [HC2H3O2] Ka =

36 Calculating pH from Ka We next set up a table…
[C2H3O2], M [H3O+], M [C2H3O2-], M Initially 0.30 Change -x +x At Equilibrium x  0.30 x We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

37 Calculating pH from Ka (x)2 (0.30) 1.8  10-5 =
Now, (x)2 (0.30) 1.8  10-5 = (1.8  10-5) (0.30) = x2 5.4  10-6 = x2 2.3  10-3 = x

38 Calculating pH from Ka pH = -log [H3O+] pH = -log (2.3  10-3)

39 Calculate the pH of a 0.50 M aqueous solution of the weak acid HF (Ka = 7.2 x 10-4).

40 Solving for pH Ka = 7.2 x 10-4 HF(aq) H+(aq) + F-(aq) Initial 0.50 M
Change -x +x Equilibrium 0.50-x x Ka = 7.2 x 10-4 See Sample Problem #5.

41 Weak Acids Using Ka to Calculate pH
Using Ka, the concentration of H+ (and hence the pH) can be calculated. Write the balanced equation showing the equilibrium. Write the equilibrium expression. Find the value for Ka. Write down the initial and equilibrium concentrations for everything except pure water. We usually assume that the change in concentration of H+ is x. Substitute into the equilibrium constant expression and solve. Remember to turn x into pH if necessary.

42 Weak Acids Using Ka to Calculate pH
Percent ionization is another method to assess acid strength. For the reaction Percent ionization relates the equilibrium H+ concentration, [H+]eqm, to the initial HA concentration, [HA]0.

43 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 [H3O+] [COO-] [HCOOH] Ka = 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.

44 Calculating Ka from the pH
pH = -log [H3O+] 2.38 = -log [H3O+] -2.38 = log [H3O+] = 10log [H3O+] = [H3O+] 4.2  10-3 = [H3O+] = [HCOO-]

45 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  10-3 = = 0.10 4.2  10-3

46 Calculating Ka from pH [4.2  10-3] [4.2  10-3] Ka = [0.10]
= 1.8  10-4

47 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% See sample problem #6.

48 Bases react with water to produce hydroxide ion.
Weak Bases Bases react with water to produce hydroxide ion.

49 Weak Bases [HB+] [OH-] [B-] Kb =
The equilibrium constant expression for this reaction is [HB+] [OH-] [B-] Kb = where Kb is the base-dissociation constant.

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

51 pH of Basic Solutions [NH4+] [OH-] [NH3] Kb = = 1.8  10-5
What is the pH of a 0.15 M solution of NH3? NH3 (aq) + H2O (l) NH4+ (aq) + OH- (aq) [NH4+] [OH-] [NH3] Kb = = 1.8  10-5

52 pH of Basic Solutions Tabulate the data. [NH3], M [NH4+], M [OH-], M
Initially 0.15 At Equilibrium x  0.15 x

53 pH of Basic Solutions (x)2 (0.15) 1.8  10-5 =

54 pH of Basic Solutions Therefore, [OH-] = 1.6  10-3 M
pOH = -log (1.6  10-3) pOH = 2.80 pH = pH = 11.20 See sample problems 8 & 9.

55 Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.

56 Weak Acids Polyprotic Acids
Let’s review using sample problems 10, 11 & 12. Polyprotic Acids Polyprotic acids have more than one ionizable proton. The protons are removed in steps not all at once: It is always easier to remove the first proton in a polyprotic acid than the second. Therefore, Ka1 > Ka2 > Ka3 etc. Most H+(aq) at equilibrium usually comes from the first ionization (i.e. the Ka1 equilibrium).

57 Weak Acids Polyprotic Acids See sample problem # 13.

58 Sulfuric Acid… … is unique among the common acids.
It is a strong acid in the first dissociation step. Ka1 = LARGE It is a weak acid in the second dissociation step. Ka2 = 1.2 x 10-2 This means calculating the pH for sulfuric acid is a little different. Only in dilute sulfuric acid solutions does the second dissociation step contribute significantly to [H+]. See sample problem # 14.

59 Calculate the pH of a 1.00 M solution of H3PO4.
Ka1 = 7.5 x 10-3 Ka2 = 6.2 x 10-8 Ka3 = 4.8 x 10-13

60 Acid-Base Properties of Salt Solutions
Nearly all salts are strong electrolytes. Therefore, salts exist entirely of ions in solution. Acid-base properties of salts are a consequence of the reaction of their ions in solution. The reaction in which ions produce H+ or OH- in water is called hydrolysis. Anions from weak acids are basic. Anions from strong acids are neutral. Anions with ionizable protons (e.g. HSO4-) are amphoteric.

61 Acid-Base Properties of Salt Solutions
To determine whether a salt has acid-base properties we use: Salts derived from a strong acid and strong base are neutral (e.g. NaCl, Ca(NO3)2). Salts derived from a strong base and weak acid are basic (e.g. NaOCl, Ba(C2H3O2)2). Salts derived from a weak base and strong acid are acidic (e.g. NH4Cl, Al(NO3)3). Salts derived from a weak acid and weak base can be either acidic or basic. Equilibrium rules apply!

62 React Arrange the following 1.0 M solutions from lowest to highest pH. Justify your answer. HBr NaOH NH4Cl NaCN NH3 HCN NaCl HF See sample problems 15 & 16.

63 Lewis Acids and Bases Brønsted-Lowry acid is a proton donor.
Focusing on electrons: a Brønsted-Lowry acid can be considered as an electron pair acceptor. Lewis acid: electron pair acceptor. Lewis base: electron pair donor. Note: Lewis acids and bases do not need to contain protons. Therefore, the Lewis definition is the most general definition of acids and bases.

64 Three Models for Acids & Bases
Definition of Acid Definition of Base Arrhenius H+ producer OH- producer Bronsted-Lowry H+ donor H+ acceptor Lewis Electron pair acceptor Electron pair donor

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