CHAPTER 13: ACIDS & BASES Section 13.1 Arrhenius Acid & Bases Svante Arrhenius, Swedish chemist (1839-1927). He understood that aqueous solutions of acids and bases conduct electricity (they are electrolytes). He theorized that acids and bases must produce ions in solution. 1) Arrhenius Acid A compound that increases the concentration of hydrogen ions, H +, in aqueous solution. 1
Important In an aqueous solution, the H + (aq) ion exist as the hydronium ion, H 3 O + (aq). HCl(g) + H 2 O(l) H 3 O + (aq) + Cl - (aq) Hydronium Ion An acid that can produce 1 H + is called a Monoprotic Acids Examples: HCl, HNO 3 An acid that can produce 2 H + is called a Diprotic Acids Example: H 2 SO 4 An acid that can produce 3 H + is called a Triprotic Acids Example: H 3 PO 4 Important In an aqueous solution, the H + (aq) ions are lost one at a time for diprotic and triprotic acids. 2
Strong Acids An acid that ionizes completely in aqueous solution. (Strong electrolyte) HBr(g) + H 2 O(l) H 3 O + (aq) + Br - (aq) Weak Acids An acid that does not completely ionize in aqueous solution. (Weak electrolyte) HCN(g) + H 2 O(l) H 3 O + (aq) + CN - (aq) Reversible H 3 PO 4 (g) + H 2 O(l) H 3 O + (aq) + H 2 PO 4 - (aq) Reversible 2) Arrhenius Base A compound that ionizes to produce hydroxide ion (OH - ) in aqueous solution. 3
Strong Base A base that ionizes completely in aqueous solution. (Strong electrolyte) KOH(s) H 2O K + (aq) + OH - (aq) Weak Base A base that does not completely ionize in aqueous solution. (Weak electrolyte) NH 3 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) Reversible Amphoteric Compounds Some substances can act as both an acid (donating H atoms) or as a base (accepting H atoms). These substances are said to be AMPHOTERIC. 4
Example: Water is an amphoteric substance. Acting as an acid NH 3 (g) + H 2 O(l) Acting as a base H 3 PO 4 (g) + H 2 O(l) NH 4 + (aq) + OH - (aq) H 3 O + (aq) + H 2 PO 4 - (aq) BrØnsted-Lowry Acids/Bases Recall what we have learned: Acids produce H + Bases produce OH - We will be using the BrØnsted-Lowry Model Acid H + donor Base H + accepter For example: HA(aq) + B - (aq) A - (aq) + HB(aq) 5
Conjugate Base - Formed when H + is removed from an acid. Conjugate Acid - Formed when H + is added to a base. Amphoteric - A species that can both accept OR donate H +. O H -1 -H + H +H + H O H H O H +1 6
Acid/Base Strengths An acid is considered to be strong if it completely ionizes. Example HCl + H 2 O Cl 1- + H 3 O 1+ The reverse reaction only occurs to a very small extent: Cl 1- + H 3 O 1+ HCl + H 2 O Since both reactions occur, an equilibrium can still be established. The strength of an acid is relative when its ionization is less than 100%. When comparing two acids that both ionize 100%, how can their strengths be compared? Answer Both acids are dissolved in a solvent different than water. 7
For example: Both HI and HCl dissolve 100% in water. In another solvent, HI ionized a bit more than HCl. This makes HI a slightly stronger acid. 8
ACID BASE Strongest Acid HClO 4 1- ClO 4 Weakest Acid H 2 SO 4 1- HSO 4 HI I 1- HBr Br 1- HCl Cl 1- HNO 3 1- NO 3 H 3 O 1+ H 2 O HSO 4 1- SO 4 2- H 2 SO 3 1- HSO 3 H 3 PO 4 1- H 2 PO 4 HNO 2 1- NO 2 HF F 1- HC 2 H 3 O 2 1- C 2 H 3 O 2 3+ Al(H 2 O) 6 A1(H 2 O) 5 (OH) 2+ H 2 CO 3 1- HCO 3 H 2 S HS 1- HClO ClO 1- HBrO BrO 1-1+ NH 4 NH 3 HCN CN 1- HCO 3 1- CO 3 2- H 2 O 2 1- HO 2 HS 1- S 2- H 2 O OH 1- Weakest Base Strongest Base 9
Molecular Structure & Acid Strength The strength of an acid depends on the ease of breaking the H-X bond to form H 1+. Two factors determine the strength: 1) The polarizability of the H-X bond. More polar easier to remove H 1+ 2) The strength of the bond is size dependent. Larger the atom The weaker the bond The stronger the acid Going down a column on the periodic table: weaker bond Stronger Acid 10
Going across a period on the periodic table: As electronegativity H-X is more polar Acid strength Example Acid strength: HF>H 2 O *** The change in the radius does not have a significant across a period.*** 11
Oxyacid Strength Structure: H-O-Y The bond polarity is the most important factor. The electronegativity of Y is very important. Example HOCl > HOBr > HOI Why? -- E.N. of Cl>Br>I For a series of oxyacids, acids strength as the # of oxygen bond to Y goes Example: Acid Strength HClO 4 >HClO 3 >HClO 2 >HClO 12
For Polyprotic Acids, acids strength as its negative charge Example: Acid Strength H 2 SO 4 >HSO 4 1-13
Section 13.2 Self-Ionization of Water The Acid-Base properties of an aqueous solution are dependent upon the equilibrium that involves H 2 O. H 2 O(l) + H 2 O(l) H 3 O + (aq) + OH - (aq) A simpler way to consider the process is the self-dissociation of a single H 2 O molecule: H 2 O(l) H + (aq) + OH - (aq) The equilibrium constant for this process is: K w =[H + ][OH - ] = 1.0 x 10-14 In pure H 2 O, [H + ] = [OH - ] If x = [H + ] = [OH - ], then, (x)(x) = 1.0 x 10-14 x 2 = 1.0 x 10-14 x = 1.0 x 10-7 = [H + ] = [OH - ] at 25 C *The solution is NEUTRAL* 14
DETERMINING CONCENTRATION The [H + ] can be determined if [OH - ] is known. The [OH - ] can also be determined if [H + ] is known. Example #1: What is the [H + ] if the [OH - ] = 2.7 x 10-4 M? Answer: Recall, K w = [H + ][OH - ] = 1.0 x 10-14 H K OH w 1.0 x10 2.7 x10 14 4 3.7 10 11 15
Example #2: What is the [OH - ] if 7.6 x 10-2 M HNO 3 is added to pure H 2 O? Answer: 7.6 x 10-2 M HNO 3 = 7.6 x 10-2 M H + Recall, K w = [H + ][OH - ] = 1.0 x 10-14 OH K H w 1.0 x10 7.6 x10 14 2 1.3 10 13 Example #3: What is the [OH - ] if.16 M H 2 SO 4 is added to pure H 2 O? Answer:.16 M H 2 SO 4 =.32 M [H + ] K w = [H + ][OH - ] = 1.0 x 10-14 OH K H w 1.0 x10.32 14 3.1 10 14 16
Most solutions are NOT neutral. However, this is still true: [H + ][OH - ] = 1.0 x 10-14 So, If [H + ] is high & [OH - ] is low ACIDIC If [H + ] is low & [OH - ] is high BASIC 17
Section 13.3 ph and poh Acidity and Basicity can be described in terms of [H + ] In 1909, SØren SØrensen proposed an alternative measure of acidity: ph is "power of the hydrogen ion" ph = -log 10 [H + ] = -log 10 [H 3 O + ] ph ranges from 0-14. Hydronium Ion A solution with ph > 7 is basic. A solution with ph < 7 is acidic A solution with ph = 7 is neutral. [H + ] = 10 -ph poh = - log 10 [OH - ] ph + poh = 14 18
ph of Strong Acids & Strong Bases Recall the strong acids: HCl, HBr, HI, HClO 4, HNO 3, and H 2 SO 4 Recall the strong bases: LiOH, NaOH, KOH, Ca(OH) 2, Sr(OH) 2, Ba(OH) 2 They dissociate 100% in H 2 O. Example, HCl(aq) + H 2 O(l) H 3 O + (aq) + Cl - (aq) If one has 1.0 M HCl, then [H + ] = 1.0 M If one has 1.0 M NaOH, then [OH - ] = 1.0 M 19
Section 13.4 Weak Acids & Acid Dissociation Constants (K a ) According to BrØnsted-Lowry Theory, this process occurs to a very small extent for a weak acid. HA(aq) + H 2 O(aq) H 3 O + (aq) + A - (aq) Weak Acid Categories 1) Molecule with ionizable H-atom HNO 2 (aq) + H 2 O(aq) H 3 O + (aq) + NO 2 - (aq) Remember that HNO 3, HBr, H 2 SO 4, HClO 4, HI, and HCl are not included here because they are strong acids!!! 2) Cation NH 4 + (aq)+ H 2 O(aq) H 3 O + (aq)+ NH 3 (aq) 20
3) Metal Cation Al 3+ (aq) Al(H 2 O) 3+ 6 (aq) - Octahedral - Hydrated Al(H 2 O) 6 3+ (aq) + H 2 O(l) H 3 O + (aq) + Al(H 2 O) 5 (OH) 2+ (aq) The Equilibrium Constant for Weak Acid HA(aq) + H 2 O(l) H 3 O + (aq) + A - (aq) HA(aq) H + (aq) + A - (aq) The equilibrium constant for this reaction is: K a H A HA The equilibrium constant for the acid dissociating is also known as the Dissociation Constant (K a ) The larger the value of K a Stronger Acid 21
Similar to the relationship of ph and [H + ], K a and pk a are related to each other in the following way: pk a = -log 10 (K a ) Weaker Acid Smaller K a Larger pk a Example The ph of a 0.115 M solution of chloroacetic acid, ClCH 2 COOH, is 1.92. Calculate K a for this monoprotic acid. Answer: ph=1.92 [H + ] = 1 x 10-1.92 [H + ] =.012 M HA(aq) H + (aq) + A - (aq) I.115 0 0 C -.012 +.012 +.012 E.103.012.012 2 H A 3 K a HA (.012).103 22 1.4 x10
What would the value of pk a be for this acid? pk a = -log 10 (1.4 x 10-3 ) = 2.9 Percent Dissociation % dissociation H HA 0 x100 What would be the % dissociation in the previous example? [H + ] = 0.012 M [HA] o = 0.115 M 0.012 % dissociation x100 0.115 10% [H + ] of a Monoprotic Acid in H 2 O. Example 23
Calculate the [H + ] in 0.10 HOCl, hypochlorous acid. K a = 3.5 x 10-8 HOCl(aq) OCl - (aq) + H + (aq) I 0.10 0 0 C -x +x +x E 0.10-x x x 2 8 x 3.5x10 0.10 x x 2 = (3.5 x 10-8 )(0.10-x) x 2 = 3.5 x 10-9 -3.5 x 10-8 x x 2 + 3.5 x 10-8 x - 3.5 x 10-9 = 0 Using the Quadratic Formula, x = -6.0 x 10-5 and 6.0 x 10-5 Therefore, [H + ] = 6.0 x 10-5 M Successive Approximations In most cases, K a is not known better than 5%. 24
For K a x a x x Where a = [HA] O x = [H + ] Situation #1: x If a x 100 5%, than we can make the assumption that x -a a. Generally, this assumption can be made if there is at least a difference of 10 3 between K a and the [HA] o. Situation #2: 25
x If a x 100% > 5%, than this assumption cannot be made. There are 2 methods that can be used to solve these types of problems: 1) Quadratic Formula 2) Successive Approximation where (a-x) is substituted with a - a x This is repeated until the value of a x stops changing. ***Generally, #1 is easier when we use graphing calculators.*** 26
Dissociation of Polyprotic Acids Acids with more than one ionizable H-atom. These acids dissociate in steps. For example, H 2 A(aq) H + (aq) + HA - (aq) K a1 HA - (aq) H + (aq) + A -2 (aq) K a2 In general, a) The anion in one-step (e.g. HA - ) dissociate in the next step. b) The dissociation constant becomes smaller with each successive step. K a1 > K a2 > K a3 *The acids become weaker with each step. *Typically, the dissociation constants decrease by at least a factor 100. 27
*Essentially, the H + in solution comes from the first dissociation. Example Calculate the ph of a 0.0010 M solution of H 2 CO 3. H 2 CO 3 (aq) HCO 3 1- (aq) H + (aq) + HCO 3 1- (aq) K a1 =4.4 x 10-7 H + (aq) + CO 3 2- (aq) K a2 =4.7 x 10-11 *Since K a1 >>K a2, essentially all of the H + comes from the first reaction. H 2 CO 3 (aq) H + (aq) + HCO 1-3 (aq) I 0.0010 0 0 C -x +x +x E 0.0010-x X x K a x x 1 4.4x10 0.0010 x Making the approximation of 0.0010 - x 0.0010 28 7
x x 0.0010 4.4x10 7 4.4 x 10-10 = x 2 2.1 x 10-5 M = x = [H + ] ph = 4.7 Is this approximation justified??? 2.1x10 0.0010 5 x 100 2.1% YES! 29
Section 13.5 Weak Bases and Their Dissociation Constants As with weak acids, weak bases fall into 2 categories: 1) Molecules NH 3 (aq) + H 2 O(aq) NH 4 + (aq) + OH - (aq) *Happens to a small extent it s weak! 2) Anions - Any anion derived from a weak acid is itself a weak base. F - (aq) + H 2 O(aq) HF(aq) + OH - (aq) For both: a) weak bases accept H +. b) H 2 O acts as a BrØnsted-Lowry acid. 30
The Equilibrium Constant for Weak Base For, B - (aq) + H 2 O(l) HB(aq) + OH - (aq) Base Dissociation Constant K b HB OH B Example Calculate the ph of a 0.25 M NaC 2 H 3 O 2. (K b of C 2 H 3 O 2- = 5.6 x 10-10 ) H 2 O(l) + C 2 H 3 O 2 1- (aq) OH - (aq) + HC 2 H 3 O 2 (aq) H 2 O 1- + C 2 H 3 O 2 OH - 1- + HC 2 H 3 O 2 I 0.25 0 0 C -x +x +x E 0.25-x x x K b HC 2 H O C 2 3 3 2 H O 2 OH 31
5.6x10 10 x 0.25 x x Assuming 0.25-x 0.25 5.6x10 10 x x 0.25 1.4 x 10-10 = x 2 1.2 x 10-5 M = x = [OH - ] * Since 1.2 x 10-5 <<<.25, the assumption of 0.25-x 0.25 is justified. poh = -log 10 [OH - ] poh = -log 10 [1.2 x 10-5 ] poh = 4.9 Since, ph + poh = 14, ph = 14.0 - ph = 4.0-4.9 = 9.1 32
The Relationships between K a and K b If the following 2 rxns. are combined: HA(aq) H + (aq) + A - (aq) K a A - (aq) + H 2 O(l) HA(aq) + OH - (aq) K b H 2 O(l) H + (aq) + OH - (aq) K w So, K w = K a x K b = 1.0 x 10-14 K a and K b are inversely related. The stronger the acid, the weaker the conjugate base. Acid/Base Properties of Salt Solutions Salt - An ionic compound with a *cation (other than H + ) *anion (other than OH - or O 2- ) To predict properties: 1) Consider the effect on ph of H 2 O 2) Combined effect of both cation/anion together 33
Cations: Weak Acid of Spectator NH + 4 (aq) + H 2 O(l) NH 3 (aq) + H 3 O + (aq) When NH + 4 (aq) reacts with water, a) The [H 3 O + ] b) The ph Zn(H 2 O) 2+ 4 (aq) + H 2 O(l) Zn(H 2 O) 3 (OH) + (aq) + H 3 O + (aq) When Zn(H 2 O) 2+ 4 (aq) reacts with water, a) The [H 3 O + ] b) The ph The ph by the addition of: 1) Transition metals 2) Main group cation, Al 3+ and Mg 2+ Spectators are: 1) Alkali ions. 2) Heavier alkaline earth cations. 34
Anions: Weak Bases of Spectators React with H 2 O to produce OH - *except for anions from strong acids. NO 2 - (aq) + H 2 O(l) HNO 2 (aq) + OH - (aq) When NO 2 - (aq) reacts with water, a) The [OH - ] b) The poh c) The ph Let s try to predict the effect on the ph of pure water by adding the following aqueous solution to pure water. Example: a) NH 4 I? NH 4 + (acidic) I - (spectator) b) KClO 4? K + (spectator) ClO 4- (spectator) 35
If both affect ph, one must determine which ion has the stronger effect. How??? If K a > K b If K b > K a Acidic Basic 36
The Common Ion Effect Describes the shift in an ionic equilibrium caused by the addition of a solute that involves an ion that takes part in the equilibrium. Consider the following equilibrium: CH 3 CO 2 H(aq) + H 2 O(aq) CH 3 CO 2 1- (aq) + H 3 O + (aq) If NaCH 3 CO 2 is added to this the solution, it dissolves completely: NaCH 3 CO 2 (aq) Na + (aq) + CH 3 CO 2 1- (aq) Both CH 3 CO 2 H and NaCH 3 CO 2 are sources of CH 3 CO 2 1-. As a result, Le Chatlier's Principle dictates that the reaction should shift to the left. The [H 3 O + ] decreases ph 37
Example Calculate [H 3 O 1+ ] and the ph of a solution that is 0.10 M in CH 3 CO 2 H and 0.20 M in NaCH 3 CO 2. K a = 1.8 x 10-5 H 2 O(l) + CH 3 CO 2 H(aq) CH 3 CO 2 1- (aq) + H 3 O + (aq) I 0.10.20 0 C -x +x +x E 0.10-x 0.20+x x K a C 2 H 3 O HC 2 2 H 3 H O 3 2 O 1 Assuming that, 0.20 + x.20 0.10-x.10 1.8x10 5 1.8x10 0.20 0.10 0.20 0.10 x x x 5 x x = 9.0 x 10-6 = [H 3 O 1+ ] ph = 5.05 38