1 The Chemistry of Acids and Bases Chapter 17

2 Acid and Bases

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5 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O(liq) ---> H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.

6 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Strong and Weak Acids/Bases

7 Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Strong and Weak Acids/Bases

8 Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH(aq) ---> Na + (aq) + OH - (aq) NaOH(aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO

9 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O(liq) e NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases

10 ACID-BASE THEORIES The most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theoryThe most general theory for common aqueous acids and bases is the BRØNSTED - LOWRY theory ACIDS DONATE H + IONSACIDS DONATE H + IONS BASES ACCEPT H + IONSBASES ACCEPT H + IONS

11 The Brønsted definition means NH 3 is a BASE in water — and water is itself an ACID ACID-BASE THEORIES

12 NH 3 is a BASE in water — and water is itself an ACID NH 3 / NH 4 + is a conjugate pair — related by the gain or loss of H + Every acid has a conjugate base - and vice-versa.

13 Conjugate Pairs

14 More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for autoion = K w K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C

15 More About Water K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x M Autoionization

16 Calculating [H 3 O + ] & [OH - ] You add mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e HO + (aq) + OH - (aq) 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) Le Chatelier predicts equilibrium shifts to the ____________. [HO + ] < at equilibrium. [H 3 O + ] < at equilibrium. Set up a ICE table.

17 Calculating [H 3 O + ] & [OH - ] You add mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) initial initial change+x+x change+x+x equilibx x equilibx x K w = (x) ( x) Because x << M, assume [OH - ] = M [H 3 O + ] = K w / = 1.0 x M

18 Calculating [H 3 O + ] & [OH - ] You add mol of NaOH to 1.0 L of pure water. Calculate [HO + ] and [OH - ]. You add mol of NaOH to 1.0 L of pure water. Calculate [H 3 O + ] and [OH - ].Solution 2 H 2 O(liq) e HO + (aq) + OH - (aq) 2 H 2 O(liq) e H 3 O + (aq) + OH - (aq) [HO + ] = K w / = 1.0 x M [H 3 O + ] = K w / = 1.0 x M This solution is _________ because [H 3 O + ] < [OH - ]

19 [H 3 O + ], [OH - ] and pH A common way to express acidity and basicity is with pH pH = log (1/ [H 3 O + ]) = - log [H 3 O + ] In a neutral solution, [HO + ] = [OH - ] = 1.00 x at 25 o C In a neutral solution, [H 3 O + ] = [OH - ] = 1.00 x at 25 o C pH = -log (1.00 x ) = - (-7) = 7

20 [H 3 O + ], [OH - ] and pH What is the pH of the M NaOH solution? [HO + ] = 1.0 x M [H 3 O + ] = 1.0 x M pH = - log (1.0 x ) = General conclusion — Basic solution pH > 7 Basic solution pH > 7 Neutral pH = 7 Neutral pH = 7 Acidic solutionpH < 7 Acidic solutionpH < 7

21 Figure 17.1

22 [H 3 O + ], [OH - ] and pH If the pH of Coke is 3.12, it is ____________. Because pH = - log [HO + ] then Because pH = - log [H 3 O + ] then log [HO + ] = - pH log [H 3 O + ] = - pH Take antilog and get [HO + ] = 10 -pH [H 3 O + ] = 10 -pH [HO + ] = = 7.6 x M [H 3 O + ] = = 7.6 x M

23 pH of Common Substances

24 Other pX Scales In generalpX = -log X and so pOH = - log [OH - ] K w = [HO + ] [OH - ] = 1.00 x at 25 o C K w = [H 3 O + ] [OH - ] = 1.00 x at 25 o C Take the log of both sides -log ( ) = - log [HO + ] + (-log [OH - ]) -log ( ) = - log [H 3 O + ] + (-log [OH - ]) pK w = 14 = pH + pOH pK w = 14 = pH + pOH

25 Equilibria Involving Weak Acids and Bases Aspirin is a good example of a weak acid, K a = 3.2 x 10 -4

26 Equilibria Involving Weak Acids and Bases AcidConjugate Base AcidConjugate Base acetic, CH 3 CO 2 HCH 3 CO 2 -, acetate ammonium, NH 4 + NH 3, ammonia bicarbonate, HCO 3 - CO 3 2-, carbonate A weak acid (or base) is one that ionizes to a VERY small extent (< 5%).

27 Equilibria Involving Weak Acids and Bases Consider acetic acid, CH 3 CO 2 H (HOAc) HOAc + H 2 O e H 3 O + + OAc - Acid Conj. base (K is designated K a for ACID) Because [H 3 O + ] and [OAc - ] are SMALL, K a << 1.

28 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7

29 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of

30 Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength

31 Relation of K a, K b, [H 3 O + ] and pH

32 K and Acid-Base Reactions Reactions always go from the stronger A-B pair (larger K) to the weaker A-B pair (smaller K). ACIDSCONJUGATE BASES STRONGSTRONG weakweak weakweak STRONGSTRONG

33 K and Acid-Base Reactions A strong acid is 100% dissociated. Therefore, a STRONG ACID —a good H + donor— must have a WEAK CONJUGATE BASE —a poor H + acceptor. HNO 3 (aq) + H 2 O(liq) e H 3 O + (aq) + NO 3 - (aq) HNO 3 (aq) + H 2 O(liq) e H 3 O + (aq) + NO 3 - (aq) STRONG A base acid weak B STRONG A base acid weak B Every A-B reaction has two acids and two bases.Every A-B reaction has two acids and two bases. Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair. Here K is very large.Here K is very large. Every A-B reaction has two acids and two bases.Every A-B reaction has two acids and two bases. Equilibrium always lies toward the weaker pair.Equilibrium always lies toward the weaker pair. Here K is very large.Here K is very large.

34 K and Acid-Base Reactions We know from experiment that HNO 3 is a strong acid. 1.It is a stronger acid than H 3 O + 2.H 2 O is a stronger base than NO K for this reaction is large

35 K and Acid-Base Reactions Acetic acid is only 0.42% ionized when [HOAc] = 1.0 M. It is a WEAK ACID HOAc + H 2 O e H 3 O + + OAc - WEAK A base acid STRONG B Because [H 3 O + ] is small, this must mean 1.H 3 O + is a stronger acid than HOAc 2.OAc - is a stronger base than H 2 O 3.K for this reaction is small

36 Types of Acid/Base Reactions Strong acid + Strong base H + + Cl - + Na + + OH - e H 2 O + Na + + Cl - Net ionic equation H + (aq) + OH - (aq) e H 2 O(liq) K = 1/K w = 1 x Mixing equal molar quantities of a strong acid and strong base produces a neutral solution.

37 Types of Acid/Base Reactions Weak acid + Strong base CH 3 CO 2 H + OH - e H 2 O + CH 3 CO 2 - This is the reverse of the reaction of CH 3 CO 2 - (conjugate base) with H 2 O. OH - stronger base than CH 3 CO 2 - K = 1/K b = 5.6 x 10 4 Mixing equal molar quantities of a weak acid and strong base produces the acid’s conjugate base. The solution is basic.

38 Types of Acid/Base Reactions Strong acid + Weak base H 3 O + + NH 3 e H 2 O + NH 4 + This is the reverse of the reaction of NH 4 + (conjugate acid of NH 3 ) with H 2 O. H 3 O + stronger acid than NH 4 + K = 1/K a = 5.6 x 10 4 Mixing equal molar quantities of a strong acid and weak base produces the bases’s conjugate acid. The solution is acid.

39 Types of Acid/Base Reactions Weak acid + Weak base Product cation = conjugate acid of weak base.Product cation = conjugate acid of weak base. Product anion = conjugate base of weak acid.Product anion = conjugate base of weak acid. pH of solution depends on relative strengths of cation and anion.pH of solution depends on relative strengths of cation and anion.

40 Types of Acid/Base Reactions: Summary

41 Calculations with Equilibrium Constants pH of an acetic acid solution. What are your observations? pH of an acetic acid solution. What are your observations? M M 0.06 M 2.0 M a pH meter, Screen 17.9

42 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib

43 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] initial change-x+x+x equilib1.00-xxx Note that we neglect [H 3 O + ] from H 2 O.

44 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula or method of approximations (see Appendix A).

45 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.

46 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = [K a 1.00] 1/2 x = [ H 3 O + ] = [ OAc - ] = 4.2 x M pH = - log [ H 3 O + ] = -log (4.2 x ) = 2.37

47 Equilibria Involving A Weak Acid Consider the approximate expression For many weak acids [H 3 O + ] = [conj. base] = [K a C o ] 1/2 [H 3 O + ] = [conj. base] = [K a C o ] 1/2 where C 0 = initial conc. of acid Useful Rule of Thumb: If 100K a < C o, then [H 3 O + ] = [K a C o ] 1/2 If 100K a < C o, then [H 3 O + ] = [K a C o ] 1/2

48 Equilibria Involving A Weak Acid Calculate the pH of a M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O e HCO H 3 O + HCO 2 H + H 2 O e HCO H 3 O + K a = 1.8 x Approximate solution [H 3 O + ] = [K a C o ] 1/2 = 4.2 x M, pH = 3.37 [H 3 O + ] = [K a C o ] 1/2 = 4.2 x M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x M [H 3 O + ] = [HCO 2 - ] = 3.4 x M [HCO 2 H] = x = M [HCO 2 H] = x = M pH = 3.47 pH = 3.47

49 Weak Bases

50 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O e NH OH - NH 3 + H 2 O e NH OH - K b = 1.8 x Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib

51 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O e NH OH - NH 3 + H 2 O e NH OH - K b = 1.8 x Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ] initial change-x+x+x equilib xx x

52 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O e NH OH - NH 3 + H 2 O e NH OH - K b = 1.8 x Step 2. Solve the equilibrium expression Assume x is small (100K b < C o ), so x = [OH - ] = [NH 4 + ] = 4.2 x M x = [OH - ] = [NH 4 + ] = 4.2 x M and [NH 3 ] = x ≈ M The approximation is valid!

53 Equilibria Involving A Weak Base You have M NH 3. Calc. the pH. NH 3 + H 2 O e NH OH - NH 3 + H 2 O e NH OH - K b = 1.8 x Step 3. Calculate pH [OH - ] = 4.2 x M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63

54 MX + H 2 O ----> acidic or basic solution? Consider NH 4 Cl NH 4 Cl(aq) ----> NH 4 + (aq) + Cl - (aq) (a)Reaction of Cl - with H 2 O Cl - + H 2 O ---->HCl + OH - Cl - + H 2 O ---->HCl + OH - baseacidacidbase baseacidacidbase Cl - ion is a VERY weak base because its conjugate acid is strong. Therefore, Cl > neutral solution Acid-Base Properties of Salts

55 NH 4 Cl(aq) ----> NH 4 + (aq) + Cl - (aq) (b)Reaction of NH 4 + with H 2 O NH H 2 O ---->NH 3 + H 3 O + NH H 2 O ---->NH 3 + H 3 O + acidbasebaseacid acidbasebaseacid NH 4 + ion is a moderate acid because its conjugate base is weak. Therefore, NH > acidic solution See TABLE 17.4 for a summary of acid-base properties of ions. Acid-Base Properties of Salts

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57 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O e HCO OH - baseacid acidbase K b = 2.1 x K b = 2.1 x Step 1.Set up concentration table [CO 3 2- ][HCO 3 - ][OH - ] initial [CO 3 2- ][HCO 3 - ][OH - ] initial change equilib equilib Acid-Base Properties of Salts

58 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O e HCO OH - baseacid acidbase K b = 2.1 x K b = 2.1 x Step 1.Set up ICE table [CO 3 2- ][HCO 3 - ][OH - ] initial [CO 3 2- ][HCO 3 - ][OH - ] initial change-x+x+x equilib xxx equilib xxx Acid-Base Properties of Salts

59 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O e HCO 3 - +OH - baseacid acidbase K b = 2.1 x K b = 2.1 x Acid-Base Properties of Salts Assume x ≈ 0.10, because 100K b < C o x = [HCO 3 - ] = [OH - ] = M Step 2.Solve the equilibrium expression

60 Calculate the pH of a 0.10 M solution of Na 2 CO 3. Na + + H 2 O ---> neutral CO H 2 O e HCO OH - baseacid acidbase K b = 2.1 x K b = 2.1 x Acid-Base Properties of Salts Step 3.Calculate the pH [OH - ] = M pOH = - log [OH - ] = 2.34 pH + pOH = 14, so pH = 11.66, and the solution is ________.