g. Looking at the equation, one can conclude that H 2 O has accepted a proton from HONH 3 HONH 3

Transcription

1 Chapter 14 Acids and Bases I. Bronsted Lowry Acids and Bases a. According to Brønsted- Lowry, an acid is a proton donor and a base is a proton acceptor. Therefore, in an acid- base reaction, a proton (H + ) is transferred from an acid to a base. b. Example: HCl (aq) + H 2 O (l) H 3 O + (aq) + Cl - (aq) In this reaction: HCl is an acid H 2 O is a base H 3 O + is the conjugate acid Cl - is the conjugate base c. Conjugate acid:the species formed when a proton is added to a base is called the conjugate acid. d. Conjugate base:the species formed when a proton is removed from an acid is called the conjugate base. e. The conjugate acid- base pair consists of the two substances related to each other by the donation and acceptance of a single proton f. Example: In the following aqueous reaction, identify the acid, the base, the conjugate acid and the conjugate base. H 2 O (l) + HONH 3 + (aq) HONH 2 (aq) + H 3 O + (aq) g. Looking at the equation, one can conclude that H 2 O has accepted a proton from HONH 3 + to form H 3 O +. HONH 3 + is a proton donor (the acid). H 2 O is the proton acceptor (the base). H 3 O + is the species formed when the proton is added to the base (the conjugate acid). HONH 2 is the species formed when a proton is removed from the acid (the conjugate base). II. The Autoionization of Water a. Pure liquid water has a very low electrical conductivity. This is because water is ionized to a very small extent. H 2 O (l) + H 2 O (l) H 3 O + (aq) + OH - (aq) b. In this reaction, H 2 O behaves as an acid as well as a base. Substances acting as either an acid or a base are said to be amphoteric. c. The above reaction is called the autoionization of water. The equilibrium constant for the autoionization of water is written as:

2 d. H 2 O is the solvent and its concentration (~ 56 M) is much larger than that of any of the solutes. Thus, [H 2 O] can be considered constant. We can therefore define a new equilibrium constant, K w, called the ion product constant. i. K w = [H 3 O + ][OH - ] ii. In pure water [H 3 O + ] = [OH - ] iii. At 25 o C, [H 3 O + ] = [OH - ] = 1.0 x 10-7 mol.l - 1 e. Thus, K w = [H 3 O + ] [OH - ] = 1.0 x10-14 at 25 o C f. The value of K w increases slightly with temperature. g. Recall: The magnitude of an equilibrium constant often depends on temperature. III. Amphoteric Substances a. A substance that can behave either as an acid or a base is called an amphoteric substance. b. Example: Consider the formation of an aqueous solution of H 2 SO 4. c. H 2 O (l) + H 2 SO 4 (aq) HSO 4 - (aq) + H 3 O + (aq) d. In this reaction H 2 O has accepted a proton from H 2 SO 4 to form H 3 O +. i. H 2 SO 4 is an acid since it donates a proton. ii. H 2 O is a base since it accepts a proton. iii. H 3 O + is the the conjugate acid since it forms when a proton is added to the base. iv. HSO 4 - is the conjugate base of the acid H 2 SO 4. e. But, HSO 4 - can also donate a proton according to the reaction: HSO 4 - (aq) + H 2 O (l) H 3 O + (aq) + SO 4-2 (aq) f. Thus, according to this reaction, HSO 4 - is an acid. HSO 4 - can behave as an acid and a base, so it is an amphoteric substance. i. HSO 4 - is the acid because it donates a proton. ii. H 2 O is the base because it accepts a proton. iii. H 3 O + is the conjugate acid, because it is formed when a proton is added to the base. iv. SO 4-2 is the conjugate base, because it is formed when a proton is removed from the acid. v. Since, H 2 SO 4 can donate two protons, it is called a diprotic acid.

3 IV. Calculations of Ion Concentrations a. At 25 o C, K w = [H 3 O + ] [OH - ] = 1.0 x b. Example: Calculate [H 3 O + ] and [OH - ] in a M HCl aqueous solution. i. HCl is a strong acid. A strong acid is defined as one that is completely dissociated in water. ii. This means that: [H 3 O + ] = [Cl - ] = mol.l - 1 iii. We know that: [H 3 O + ] [OH - ] = 1.0 x iv. (0.050) x [OH - ] = 1.0 x v. [OH - ] = 2.0 x mol.l - 1 vi. [H 3 O + ] = mol.l - 1 and [OH - ] = 2.0 x mol.l - 1 c. Hence, in a M HCl solution: [H 3 O + ] >> [OH - ] V. The ph Scale a. The ph scale provides a convenient way to express the acidity of dilute aqueous solutions. The ph is calculated using a decimal logarithm or log base 10. ph = - log [H + ] or ph = - log [H 3 O + ] b. Remember that you can write [H + ] = [H 3 O + ] c. Thus, [H + ] = 10 - ph d. Similarly, poh = - log [OH - ] e. The ph of a solution is usually measured using a ph meter. A ph meter is an electronic device with an electrode that can be inserted into the solution of unknown ph f. The electrode contains an acidic aqueous solution enclosed in a special glass membrane that allows H + ions to migrate. The difference in ph between the solution in the electrode and the solution of unknown ph generates an electrical potential which is registered on the ph meter. g. We know, K w = [H + ] [OH - ] i. Take the log of both sides: log K w = log [H + ] + log [OH - ] ii. Multiply both sides by - 1: - log K w = - log [H + ] - log[oh - ] iii. Define pk w by: pk w = - log K w pk w = ph + poh iv. At 25 o C K w = 1.0 x pk w = h. Thus, ph + poh = i. A neutral solution has ph = 7. j. A solution having ph < 7 is called an acidic solution. k. A solution having ph > 7 is called a basic solution. l. Note: You should know the rule on significant figures for logarithms. m. Rule: The number of decimal places in the log is equal to the number of significant figures in the original number.

4 VI. Calculating the ph of Strong Monoprotic Acid Solutions a. There are relatively few strong acids. Consider an aqueous solution of b. 1.0 M HCl. We know that HCl is a strong acid since it dissociates completely in water. Thus, HCl (aq) H + (aq) + Cl - (aq) c. Note: This reaction reaches a chemical equilibrium, in which the HCl solution contains only an extremely small number of HCl molecules and consists mostly of H + and Cl - ions. Hence, while the reverse arrow could be shown in the above reaction, it is generally ignored. i. For strong acids, we normally do not represent this reaction with two anti- parallel arrows. ii. the equilibrium reaction for a strong acid is expressed as above, with only iii. Whenever we deal with the acid- base equilibrium it is important to understand the solution components. In a 1.0 M HCl solution, there will mostly be H + and Cl - ions. d. [H + ] = 1.0 M and [Cl - ] = 1.0 M e. ph of a 1.0 M HCl solution would be: ph = - log (1.0) f. ph = 0 g. Since 0 < 7, the solution is acidic VII. Weak Acid Solutions a. Recall the definition of a weak acid: "A weak acid does not ionize completely when dissolved in water." Consider a general weak acid, HA. According to the definition: HA (aq) H + (aq) + A - (aq) b. In solution, there are H + and A - ions and HA molecules. Thus, the ionization reaction of a weak acid is represented by forward and reverse arrows. The forward and reverse arrows indicate that the ions and molecules are at equilibrium at a given temperature. Hence, the equilibrium constant is expressed as: c. Here, K eq is the same as K c, discussed in the previous chapter. Since the reaction depicts the dissociation of a weak acid, the equilibrium constant is often expressed as K a.

5 d. Let us examine the differences between a strong and weak acid: e. Consider a strong acid, HA. i. The ionization reaction is expressed as: HA (aq) H + (aq) + A - (aq) ii. The equilibrium constant is: f. Note: The value of K a is very large, since strong acids are almost fully dissociated ([HA] 0). Hence, this expression is not often used for strong acids. g. Consider a weak acid, HA. i. The ionization reaction is expressed as: HA (aq) H + (aq) + A - (aq) ii. The equilibrium constant is: h. Note: The value of K a is very small. The smaller the K a value, the weaker the acid. i. Acids (HA) i. Strong: HA (aq) H + (aq) + A - (aq) ii. Weak: HA (aq) H + (aq) + A - (aq) j. In both equations, A - is the conjugate base. k. The stronger the acid, the weaker the conjugate base. l. The weaker the acid, the stronger the conjugate base. m. The K a values at 25 o C for some weak acids are given in the table on the DVD. Note: The smaller K a, the weaker the acid, but the stronger the conjugate base. VIII. Calculating the ph of Weak Mnoprotic Acid Solutions a. Example: Calculate the ph of 1.0 M CH 3 COOH (K a = 1.8 x 10-5 ). b. Step 1. Write the ionization reaction of the acid. CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) K a = 1.8 x 10-5 c. Step 2. Write the K a value of the acid (given in the problem). d. Step 3. Write the autoionization equation for H 2 O. H 2 O (l) H + (aq) + OH - (aq) K w = 1.0 x e. Step 4. Compare K a to K w. i. If K a > K w, we conclude that CH 3 COOH is the dominant source for H +. ii. Note: if K a < K w, we cannot ignore H + from H 2 O. In this case, the ph is f. Step 5. Assume x amount of the acid dissociates

6 g. Note: According to the stoichiometry, if x mol.l - 1 CH 3 COOH acid dissociates then, x mol.l - 1 H + and x mol.l - 1 CH 3 COO - are formed. h. Step 6. The equilibrium mole fractions are: i. Step 7. Write the equilibrium constant expression and substitute the respective values. j. Step 8. Assumption: Since the K a value for the acid is small, assume only a very small amount of the acid ionizes. Therefore, one can assume x to be negligible in (1.0 - x). Thus, the equilibrium constant expression becomes k. Step 9. Examine the equilibrium concentrations [H + ] = x = 4.2 x 10-3 mol.l - 1 ph = - log [H + ] ph = - log (4.2 x 10-3 ) ph = 2.37

7 l. Typically, K a values for acids are known with an accuracy of + 5%. Hence, one can check the validity of the assumption made in Step 8. The concentration of x we calculated is 4.2 x10-3 mol.l - 1 and the initial concentration of the acid is 1.0 M. m % < 5 % n. Therefore, the assumption is valid. If this assumption was not valid, then x should be obtained by solving the quadratic equation in step 8. IX. Calculating the ph of Strong Base Solutions a. By definition, a strong base is completely dissociated in water. Therefore, the formation of a 0.10 M NaOH solution is expressed as: b. The concentrations of Na + and OH - ions are 0.10 M. c. Recall: ph = - log [H + ] and ph = - log [OH - ] d. Hence, the ph of a 0.10 M NaOH solution is calculated as: poh = - log (0.10) poh = 1.00 ph + poh = ph = X. Weak Base Solutions a. By definition, a weak base does not dissociate completely when dissolved in water. When a weak base (B) dissolves in H 2 O, it accepts a proton from H 2 O according to the reaction: B (aq) + H 2 O (l) BH + (aq) + OH - (aq) b. Ignoring water, the expression for the equilibrium constant is written as: c. Note: [H 2 O] is assumed to be constant. Here, K b is defined as the base dissociation constant. The relationship between K b and pk b is: pk b = - log K b d. According to this equation, as the value of K b increases, the value of pk b decreases. A larger value of K b indicates a stronger base. e. In aqueous solution, the two large classes of weak bases are: i. Nitrogen- containing molecules such as NH 3 and amines. ii. The anions of weak acids. f. Remember: The weaker the acid, the stronger the conjugate base. g. Note: The statement stronger conjugate base does not mean that the conjugate base of a weak acid is strong (i.e. fully dissociated). The conjugate base of a weak acid is a weak base.

8 XI. Calculating ph of Sulfuric Acids a. Sulfuric acid, H 2 SO 4, can release two protons when dissolved in water. Hence, it is called a diprotic acid and has two equilibrium constants. The first dissociation reaction for H 2 SO 4, ignoring water, is written as: H 2 SO 4 (aq) H + (aq) + HSO - 4 (aq) K a1 = very large b. Since the K a1 value is very large, H 2 SO 4 behaves like a strong acid in the first dissociation step. c. Then, HSO - 4 (aq) H + (aq) + SO (aq) K a2 = 1.2 x 10-2 d. Note: In the second dissociation step, H 2 SO 4 behaves like a weak acid. XII. Acid- Base Properties of Salts a. Ionic compounds are called salts. i. Salts that dissolve in water break into ions. ii. These ions move about independently in dilute solutions. iii. Sometimes, these ions can behave as acids or bases. b. Remember: weak acids and bases are always associated with relatively low K a and K b values, respectively. c. Salts that dissolve in water can produce acidic, neutral or basic solutions. First, we examine salts that produce neutral solutions. XIII. Salts that Produce Neutral Solutions a. Consider the dissolution of solid sodium chloride, NaCl (s) in water. The dissociation reaction for NaCl is expressed as: NaCl (s) Na + (aq) + Cl - (aq) b. Na + (aq) is the cation of a strong base, NaOH. The dissociation reaction for NaOH is expressed as: NaOH (s) Na + (aq) + OH - (aq) c. Thus, Na + (aq) is the very weak conjugate acid of the strong base, NaOH. d. Na + (aq) has no affinity for OH - (aq) and cannot produce H + (aq). e. Thus, the formation of Na + (aq) from NaCl has no effect on the ph. f. Cl - (aq) is the anion of the strong acid, HCl. The dissociation reaction for HCl is expressed as: HCl (g) H + (aq) + Cl - (aq) g. Thus, Cl - (aq) is the very weak conjugate base of the strong acid, HCl. h. Cl - (aq) has no affinity for H + (aq) nor can it produce OH - (aq). i. Thus, the formation of Cl - (aq) from NaCl has no effect on the ph. j. Therefore, the dissolution in water of a salt consisting of a cation from a strong base and an anion from a strong acid has no effect on the concentration of H +. Since [H + ] is not affected, the ph does not change. Recall that the ph of water is 7.00 at 25 o C. Hence, these salts produce a neutral solution when dissolved in water. k. Other examples of salts that produce neutral aqueous solutions are KCl, KNO 3, NaNO 3, NaBr, NaI.

9 XIV. Salts that Produce Basic Solutions a. Consider sodium acetate, NaC 2 H 3 O 2. The dissociation reaction for NaC 2 H 3 O 2, ignoring water, is written as: NaC 2 H 3 O 2 Na + (aq) + C 2 H 3 O - 2 (aq) b. Na + (aq) is the cation of a strong base (NaOH). Na + (aq) has no affinity for OH - (aq), nor can it produce H + (aq). Thus, Na + (aq) has no effect on the ph. c. C 2 H 3 O - 2 (aq) is the anion of the weak acid, HC 2 H 3 O 2 (acetic acid). The dissociation of HC 2 H 3 O 2 is given as: HC 2 H 3 O 2 (aq) H + (aq) + C 2 H 3 O - 2 (aq) d. C 2 H 3 O - 2 (aq) is the weak conjugate base of the weak acid, HC 2 H 3 O 2. e. According to the Brønsted- Lowry definition, a base is a proton acceptor. f. C 2 H 3 O - 2 (aq) has some affinity for H + (aq). C 2 H 3 O - 2 (aq) accepts H + (aq) from H 2 O to form HC 2 H 3 O 2 by the reaction: C 2 H 3 O - 2 (aq) + H 2 O (l) HC 2 H 3 O 2 (aq) + OH - (aq) g. C 2 H 3 O - 2 (aq) has some affinity for H + (aq) resulting in the formation of OH - (aq) and thus, producing a basic solution. h. Therefore, the dissolution in water of a salt consisting of a cation from a strong base and an anion from a weak acid; results in the formation of OH - (aq). i. These salts produce a basic solution when dissolved in water. j. Other examples of salts that produce basic solutions include KNO 2, KF, KCN, etc. k. Consider ammonium chloride, NH 4 Cl. Ignoring water, the dissociation reaction for NH 4 Cl is: NH 4 Cl (s) NH + 4 (aq) + Cl - (aq) l. Cl - (aq) is the conjugate base of the strong acid, HCl. Cl - (aq) is a very weak base (weaker than H 2 O). Hence, Cl - (aq) does not affect the ph. NH + 4 (aq) is the weak conjugate acid of the weak base NH 3. An acid, by definition, is a proton donor. Thus, NH + 4 (aq) reacts with water as follows: NH + 4 (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) m. Ignoring water, the reaction can be simply written as: NH + 4 (aq) NH 3 (aq) + H + (aq) n. Formation of H + (aq) implies that an acidic solution is produced. o. Therefore, dissolution in water of salts consisting of an anion from a strong acid and a cation that is the conjugate acid of a weak base, results in the formation of an acidic solution. p. Note: In principle, all metal ions react with water to produce acidic solutions. Furthermore, the higher the charge of the metal ions (Al +3, Cr +3, Fe +3, etc), the more acidic the aqueous solution produced by the corresponding salts.

10 q. C 2 H 3 O 2 - (aq) has some affinity for H + (aq). C 2 H 3 O 2 - (aq) accepts H + (aq) from H 2 O to form HC 2 H 3 O 2 by the reaction: C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2 (aq) + OH - (aq) r. C 2 H 3 O 2 - (aq) has some affinity for H + (aq) resulting in the formation of OH - (aq) and thus, producing a basic solution. s. Therefore, the dissolution in water of a salt consisting of a cation from a strong base and an anion from a weak acid; results in the formation of OH - (aq). t. These salts produce a basic solution when dissolved in water. u. Other examples of salts that produce basic solutions include KNO 2, KF, KCN, etc. XV. Salts that Produce Acidic Solutions a. Consider ammonium chloride, NH 4 Cl. Ignoring water, the dissociation reaction for NH 4 Cl is: NH 4 Cl (s) NH 4 + (aq) + Cl - (aq) b. Cl - (aq) is the conjugate base of the strong acid, HCl. Cl - (aq) is a very weak base (weaker than H 2 O). Hence, Cl - (aq) does not affect the ph. NH 4 + (aq) is the weak conjugate acid of the weak base NH 3. An acid, by definition, is a proton donor. Thus, NH 4 + (aq) reacts with water as follows: NH 4 + (aq) + H 2 O (l) NH 3 (aq) + H 3 O + (aq) c. Ignoring water, the reaction can be simply written as: NH 4 + (aq) NH 3 (aq) + H + (aq) d. Formation of H + (aq) implies that an acidic solution is produced. e. Therefore, dissolution in water of salts consisting of an anion from a strong acid and a cation that is the conjugate acid of a weak base, results in the formation of an acidic solution. f. Note: In principle, all metal ions react with water to produce acidic solutions. Furthermore, the higher the charge of the metal ions (Al +3, Cr +3, Fe +3, etc), the more acidic the aqueous solution produced by the corresponding salts. g. C 2 H 3 O 2 - (aq) has some affinity for H + (aq). C 2 H 3 O 2 - (aq) accepts H + (aq) from H 2 O to form HC 2 H 3 O 2 by the reaction: C 2 H 3 O 2 - (aq) + H 2 O (l) HC 2 H 3 O 2 (aq) + OH - (aq) h. C 2 H 3 O 2 - (aq) has some affinity for H + (aq) resulting in the formation of OH - (aq) and thus, producing a basic solution. i. Therefore, the dissolution in water of a salt consisting of a cation from a strong base and an anion from a weak acid; results in the formation of OH - (aq). j. These salts produce a basic solution when dissolved in water. k. Other examples of salts that produce basic solutions include KNO 2, KF, KCN, etc.