2 pH of salt solutions 1.Salts derived from strong acids and strong bases These consist of cations from strong bases and the anions from strong acids Examples: NaCl, KNO 3, NaClO 4, KBr, etc. These have no effect on the [H + ] when dissolved in water, since cations of strong bases are very weak conjugate acids (do not react with water), and anions of strong acids are very weak bases that do not react with water, thus resulting in neutral solutions.

3 2. Salts derived from weak acids and strong bases These are salt with cations from strong bases (such as Na + or K +, which are very weak conjugate acids that do not react with water) and anions which are the conjugate bases of a weak acid (strong conjugate bases that react with water to produce hydroxide ions). The aqueous solution of these salts will be basic. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. 2. Salts derived from weak acids and strong bases These are salt with cations from strong bases (such as Na + or K +, which are very weak conjugate acids that do not react with water) and anions which are the conjugate bases of a weak acid (strong conjugate bases that react with water to produce hydroxide ions). The aqueous solution of these salts will be basic. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. 2. Salts derived from weak acids and strong bases These are salt with cations from strong bases (such as Na + or K +, which are very weak conjugate acids that do not react with water) and anions which are the conjugate bases of a weak acid (strong conjugate bases that react with water to produce hydroxide ions). The aqueous solution of these salts will be basic. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc. 2. Salts derived from weak acids and strong bases These are salt with cations from strong bases (such as Na + or K +, which are very weak conjugate acids that do not react with water) and anions which are the conjugate bases of a weak acid (strong conjugate bases that react with water to produce hydroxide ions). The aqueous solution of these salts will be basic. Examples: NaF, KCN, NaC 2 H 3 O 2, Na 3 PO 4, Na 2 CO 3, K 2 S, Na 2 C 2 O 4, etc.

4 Examples NO 2 - (aq) + H 2 O (l)  HNO 2(aq) + OH - (aq) CN - + H 2 O  HCN + OH - ClO - + H 2 O  HClO + OH - PO H 2 O  HPO OH - You should observe that OH - is produced which means the reaction should be basic and we need to calculate K b for the reaction.

5 Calculate the pH of a 0.45M NaCN solution. The K a value for HCN is 6.2 x Solution: Since HCN is a weak acid, the Cyanide ion must have significant affinity for protons. CN - (aq) + H 2 O (l) HCN (aq) + OH - (aq) Initial Change-X +x +x Equilibrium 0.45 – x x x K b = [HCN][OH - ] [CN - ] The value of K b can be calculated from K w and the K a value for HCN. K b = = = 1.61 x K w K a (for HCN) 1.0 x x

6 Thus: K b = 1.61 x = = = [HCN][OH - ] [CN - ] (x)(x) x ~ x x = 2.69 x ~ RE = 2.69*10 -3 /0.45}*100% = 0.6% The assumption is valid, where 0.45 >> x x = [OH - ] = 2.69 x M pOH = -log[OH - ] = 2.57 pH = – 2.57 = 11.43

7 What is the pH of a 1.0 M Na 2 CO 3 solution? For H 2 CO 3, K a1 = 4.3*10 -7 and K a2 = 4.8* Solution : In the solution, [Na + ] = 2.0 M and [CO 3 2– ] = 1.0 M. H 2 O + CO 3 2–  HCO OH – K b = K w / K a2 = 2.1*10 -4 H 2 O  H + + OH -K w = Comparing the two equilibria, it is evident that the first gives about more OH - than that produced from water. Thus, CO 3 2– is a much stronger base than water and the solution is basic.

8 H 2 O + CO 3 2–  HCO OH – Initial Change -x+x +x Equilibrium 1.0 – x x x K b = K w / K a2 = 2.1* *10 -4 = {X 2 /(1.0 – X)}, assume 1.0 >> x X = 1.45*10 -2 RE = {1.45*10 -2 /1.0} * 100% = 1.45% Assumption is OK, 1.0 >> x X = [OH - ] = 1.45*10 -2 M pOH = 1.84 pH = 14 – 1.84 = 12.16

9 3. Salts derived from weak bases and strong acids Salt containing the conjugate acid of a weak base (strong conjugate acid that dissociates to produce H + ), and the anion of a strong acid (very weak conjugate base that does not react with water and will not affect acidity, like Cl -, Br -, I -, NO 3 -, ClO 4 - ) Examples: NH 4 Cl, ArNH 3 NO 3, etc NH 4 +  NH 3 + H + ArNH 3 +  ArNH 2 + H + The solution is therefore acidic when such salts are dissolved in water.

10 Find the pH of a 0.1 M NH 4 Cl soln. K b = 1.8* NH 4 +  NH 3 + H – x x x K a = { /1.8*10 -5 } = 5.6* * = x 2 /(0.1 – x), assume 0.1>>>x X = 7.5*10 -6, RE = {7.5*10 -6 /0.1}*100= 7.5*10 -3 % (OK) X = [H + ] = 7.5*10 -6 M, pH = 5.13

11 4. Salts derived from a weak base and a weak acid Here the conjugate acid of the base will be strong and will react with water to produce H +. At the same time, the conjugate base of the weak acid will also be strong and will react with water to produce OH -. The question is: which is larger the amount of H + or OH - ?

12 Look at the equilibrium constants of the conjugate acid and the conjugate base. The acidity of the salt solution will depend on the larger one. Example: NH 4 HS Acid part: NH 4 +  NH 3 + H + K a = { /1.8*10 -5 } =5.6* Base part: HS - + H 2 O  H 2 S + OH - K b = { /k a1 } = { /9.5*10 -8 } = 1.1*10 -7 It is evident that the equilibrium constant (K b ) responsible for OH - is much greater than that related to H + production (K b ), thus a basic solution is obtained

13 5. Salts derived from a cation of small size and large charge and an anion from a strong acid The cations that have a small size but large charge are conjugate acids (Lewis acids) and will react with water to produce H +, while anions from the strong acids will not react with water as these are very weak conjugate bases. Therefore, such salts will produce acidic solutions. Examples: AlCl 3, Fe(NO 3 ) 3, etc.

14 Lewis Acids and Bases - The most general acid-base definition is based on electrons and is called the Lewis definition (same Lewis that gave us Lewis structures). Acid = Electron pair acceptor Base = Electron pair donor Lewis acid: An electron pair acceptor (e.g. AlCl 3 )

15 Lewis Acids and Bases Example:AlCl 3 + Cl -  AlCl 4 - AlCl 3 accepts a pair of electrons = Acid Cl - donates a pair of electrons = Base The above is an acid base reaction in the Lewis sense; AlCl 3 & Cl - are Lewis acids & bases but not Brønsted or Arrhenius acids & bases.

16 Lewis Acids Lewis acids are defined as electron-pair acceptors. Atoms with an empty valence orbital can be Lewis acids.

17 Lewis Bases 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.

18 Metal Cations as Lewis Acids M H 2 O (l) M(H 2 O) 4 2+ Metal ions can accept electron pairs from water molecules to form complexes. An example is nickel which forms the hexa aqua complex: Ni H 2 O (l) Ni(H 2 O) 6 2+ (aq)

19 K a Values of Some Hydrated Metal Ions at 25 o C Ion K a Fe 3+ (aq) 6 x Sn 2+ (aq) 4 x Cr 3+ (aq) 1 x Al 3+ (aq) 1 x Be 2+ (aq) 4 x Cu 2+ (aq) 3 x Pb 2+ (aq) 3 x Zn 2+ (aq) 1 x Co 2+ (aq) 2 x Ni 2+ (aq) 1 x

20 Identifying Lewis Acids and Bases Problem: Identify the Lewis acids and bases in the following reactions: (a) F - + BF 3  BF 4 - (b) Co H 2 O  Co(H 2 O) 6 2+ (c) NH 3 + H +  NH 4 + (a) The BF 3 accepted an electron pair from the fluoride ion. BF 3 is the acid and F - is the base. (b) The Co 2+ ion accepted the electron pairs from the water molecules. Co 2+ is the acid and H 2 O is the base. (c) The H + ion accepted the electron pair from the ammonia molecule. H + is the acid and water is the base.

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22 Solution: Since the Al(H 2 O) 6 3+ ion is a stronger acid than water, the dominate equilibrium will be: 1.4 x = K a = [Al(OH)(H 2 O) 5 2+ ][H + ] [Al(H 2 O) 6 3+ ] Calculate the pH of a M AlCl 3 solution. The K a value for the Al(H 2 O) 6 3+ ion is 1.4 x Al(H 2 O) 6 3+ (aq) Al(OH)(H 2 O) 5 2+ (aq) + H + (aq) 0.01 –xXX

23 Thus: K a = [Al(OH)(H 2 O) 5 2+ ][H + ] [Al(H 2 O) 6 3+ ] 1.4*10 -5 = = (x) (x) x x x = 3.7 x The approximation is valid [H + ] = x = 3.7 x M and pH = 3.43 RE = {3.7*10 -4 /0.01} * 100% = 3.7% Assume 0.01 >> x

24 6. Salts derived from a cation of small size and large charge and an anion from a weak acid The cations that have a small size but large charge are conjugate acids (Lewis acids) and will react with water to produce H +, while anions from the weak acids will react with water to produce OH -. Therefore, such acidity will depend on the values of K a (for the hydrated cation) and the K b for the anion. K a > K b results in an acidic solution K b > K a results in a basic solution K a = K b results in a neutral solution Examples: Al(CH 3 COO - ) 3, Fe(NO 2 ) 3, etc.

25 Determine whether an aqueous solution of iron(III) nitrite, Fe(NO 2 ) 3, is acidic, basic, or neutral. Plan: The formula consists of the small, highly charged, and therefore weakly acidic, Fe 3+ cation and the weakly basic NO 2 - anion of the weak acid HNO 2, To determine the relative acidity of the solution, we write equations that show the reactions of the ions with water, and then find K a and K b of the ions to see which ion reacts to form to a greater extent.

26 Fe(H 2 O) 6 3+ (aq) + H 2 O (l) Fe(H 2 O) 5 OH 2+ (aq) + H 3 O + (aq) NO 2 - (aq) + H 2 O (l) HNO 2(aq) + OH - (aq) Obtaining K a and K b of the ions: For Fe 3+ (aq) K a = 6 x For NO 2 - (aq), K b must be determined: K b of NO 2 - = = = 1.4 x K a of HNO 2 KwKw 1.0 x x Since K a of Fe 3+ > K b of NO 2 -, the solution is acidic.

27 Are the solutions of the following salts neutral, basic or acidic NaCl?NaAc?NH 4 Cl? NH 4 Ac? KClO 4 ?CH 3 NH 3 ClO 4 ?

28 Predict whether an aqueous solution of each of the following salts will be acidic, basic, or neutral. a) NH 4 C 2 H 3 O 2 b) NH 4 CN c) Al 2 (SO 4 ) 3 a)The ions are the ammonium and acetate ions, K a for NH 4 + is 5.6 x , and K b for C 2 H 3 O 2 - is 5.6 x Since they are equal, the solution will be neutral and the pH close to 7.

29 b)The solution will contain the ammonium and cyanide ions, the K a value for NH 4 + is 5.6 x , and Since K b for CN - is much larger than K a for NH 4 +, this solution will be basic. c) This solution contains the hydrated Aluminum ion and the sulfate ion. K a for Al(H 2 O) 6 3+ = 1.4 x 10 -5, for sulfate, K b = 8.3 x ; therefore this solution will be acidic. K b (for CN - ) = = 1.6 x KwKw K a (for HCN)

30 Type of Salt Examples Comment pH of Solution Cation is from Neither acts as strong base; KCl, KNO 3 an acid or a Neutral anion is from NaCl, NaNO 3 base strong acid Cation is from Anion acts as strong base; NaC 2 H 3 O 2 a base; cation Basic anion is from KCN, NaF has no effect weak acid on pH Cation is conjugate Cation acts as acid of weak base; NH 4 Cl, an acid; anion Acidic anion is from NH 4 NO 3 has no effect strong acid on pH Summary of Acidity of Salt solutions

31 Cation is conjugate Cation acts as Acidic if : acid of weak base NH 4 C 2 H 3 O 2 an acid; anion K a > K b anion is conjugate NH 4 CN acts as a base Basic if : base of weak acid K b > K a Neutral if : K a = K b Cation is highly Hydrated cation charged metal ion; Al(NO 3 ) 3, acts as an acid; Acidic anion is from FeCl 3 anion has no strong acid effect on pH Acidic: K a >K b Basic: K b >K a Neutral: K a = K b Hydrated cation acts as an acid; anion acts as a base Al 2 (SO 4 ) 3 Fe(NO 2 ) 3 Cation is highly charged metal ion; anion is conjugate base of weak acid Cation is conjugate Cation acts as Acidic if : acid of weak base NH 4 C 2 H 3 O 2 an acid; anion K a > K b anion is conjugate NH 4 CN acts as a base Basic if : base of weak acid K b > K a Neutral if : K a = K b Cation is highly Hydrated cation charged metal ion; Al(NO 3 ) 3, acts as an acid; Acidic anion is from FeCl 3 anion has no strong acid effect on pH

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33 Selected Problems 5-7, 12, 13, 15-19, 21-24, 26, 29, 31-35, 37, 41-47, 51, 52, 55-58, 61-63, 67-70, 73, 75.