16.6 Weak Acids Weak acids are only partially ionized in aqueous solution: mixture of ions and un-ionized acid in solution. Therefore, weak acids are in equilibrium: HA(aq) + H 2 O(l) H 3 O + (aq) + A (aq) Or HA(aq) H + (aq) + A (aq) We can write an equilibrium constant expression for this dissociation: [ ][ ] + H3O A Ka = or K [ HA] K a is called the acid-dissociation constant. + [ H ][ A ] [ HA] The larger the K a, the stronger the acid. K a is larger since there are more ions present at equilibrium relative to un-ionized molecules. If K a >> 1, then the acid is completely ionized and the acid is a strong acid. a = Calculating Ka from ph In order to find the value of K a, we need to know all of the equilibrium concentrations. The ph gives the equilibrium concentration of H +. - 1 -
Sample Exercise 16.10 (p. 688) A student prepared a 0.10 M solution of formic acid (HCHO 2 ) and measured its ph. The ph at 25 o C was found to be 2.38. Calculate the K a for formic acid at this temperature. (1.8 x 10-4 ) Practice Exercise 1 (16.10) A 0.50 M solution of an acid HA has ph = 2.24. What is the value of K a for the acid? a) 1.7 x 10-12 b) 3.3 x 10-5 c) 6.6 x 10-5 d) 5.8 x 10-3 e) 1.2 x 10-2 - 2 -
Practice Exercise 2 (16.10) Niacin, one of the B vitamins, has the following molecular structure: A 0.020 M solution of niacin has a ph of 3.26. What is the acid-dissociation constant, K a, for niacin? (1.6 x 10-5 ) Percent ionization is another method to assess acid strength. For the reaction % ionization = HA(aq) H + (aq) + A (aq) + [ H ] [ HA] equilibrium initial 100 The higher the percent ionization, the stronger the acid. However, we need to keep in mind that percent ionization of a weak acid decreases as the molarity of the solution increases. - 3 -
Sample Exercise 16.11 (p. 689) A 0.10 M solution of formic acid (HCOOH) contains 4.2 x 10-3 M H + (aq). Calculate the percentage of the acid that is ionized. (4.2%) Practice Exercise 1 (16.11) A 0.077 M solution of an acid HA has ph = 2.16. What is the percentage of the acid that is ionized? a) 0.090% b) 0.69% c) 0.90% d) 3.6% e) 9.0% Practice Exercise 2 (16.11) A 0.020 M solution of niacin has a ph of 3.26. Calculate the percent ionization of the niacin. (2.7%) - 4 -
Using K a to Calculate ph Using K a, we can calculate the concentration of H + (and hence the ph). 1. Write the balanced chemical equation clearly showing the equilibrium. CH 3 COOH (aq) H + (aq) + CH 3 COO - (aq) 2. Write the equilibrium expression. Look up the value for K a (in a table). K a = [H + ][ CH 3 COO - ] = 1.8 x 10-5 [CH 3 COOH] 3. Write down the initial and equilibrium concentrations for everything except pure water. We usually assume that the equilibrium concentration of H + is x. 4. Substitute into the equilibrium constant expression and solve. K a = 1.8 x 10-5 = [H + ][C 2 H 3 O - 2 ] = (x)(x) [HC 2 H 3 O 2 ] 0.30 - x Note that K a is very small (1.8 x 10-5 ) relative to [HC 2 H 3 O 2 ] (0.30 M). Keep this x x 2 = 1.8 x 10-5 0.30 x Neglect x in the denominator since it is extremely small relative to 0.30. (Use ballpark figure of 10 3 X difference for neglecting x in the denominator ) x 2 = (1.8 x 10-5 )(0.30) = 5.4 x 10-6 x 2 = (5.4 x 10-6 ) X = 2.3 x 10-3 M = [H + ] Check: Compare x with original [HC 2 H 3 O 2 ] of 0.30 M: 2.3 x 10-3 M x 100% = 0.77%, which is < 5% 0.30 M 5. Convert x ([H + ]) to ph. ph = - log (2.3 x 10-3 ) = 2.64-5 -
What do we do if we are faced with having to solve a quadratic equation in order to determine the value of x? If x is <5% of the initial concentration, the assumption is probably a good one. If x>5% of the initial concentration, then it may be best to solve the quadratic equation or use successive approximations. Sample Exercise 16.12 (p. 691) Calculate the ph of a 0.20 M solution of HCN. Refer to Table 16.2 for K a. (5.00) - 6 -
Practice Exercise 1 (16.12) What is the ph of a 0.40 M solution of benzoic acid, C 6 H 5 COOH? (The K a value for benzoic acid is given in Table 16.2.) a) 2.30 b) 2.10 c) 1.90 d) 4.20 e) 4.60-7 -
Practice Exercise 2 (16.12) The K a for niacin is 1.6 x 10-5. What is the ph of a 0.010 M solution of niacin? (3.41) - 8 -
Sample Exercise 16.13 (p. 693) Calculate the percentage of HF molecules ionized in a) a 0.10 M HF solution (7.9%) b) a 0.010 M HF solution (23%) - 9 -
Practice Exercise 1 (16.13) What is the ph of a 0.010 M solution of HF? a) 1.58 b) 2.10 c) 2.30 d) 2.58 e) 2.64 Practice Exercise 2 (16.13) In Practice Exercise 16.11, we found that the percent ionization of niacin (K a = 1.5 x 10-5 ) in a 0.020 M solution is 2.7%. Calculate the percentage of niacin molecules ionized in a solution that is a) 0.010 M (3.9%) b) a 1.0 x 10-3 M (12%) - 10 -
Polyprotic Acids Polyprotic acids have more than one ionizable proton. The protons are removed in successive steps. e.g. H 2 SO 3 (sulfurous acid): H 2 SO 3 (aq) H + (aq) + HSO 3 (aq) K a1 = 1.7 x 10 2 HSO 3 (aq) H + (aq) + SO 3 2 (aq) K a2 = 6.4 x 10 8 It is always easier to remove the first proton in a polyprotic acid than the second: K a1 > K a2 > K a3, etc. The majority of the H + (aq) at equilibrium usually comes from the first ionization (i.e., the K a1 equilibrium). If the successive K a values differ by a factor of 0 3, we can usually get a good approximation of the ph of a solution of a polyprotic acid by only considering the first ionization. 16.7 Weak Bases Weak bases remove protons from substances. There is an equilibrium between the base and the resulting ions: Weak base + H 2 O(l) conjugate acid + OH (aq) Example: NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq). The base-dissociation constant, K b, is defined as + [ ][ ] NH 4 OH K b = [ NH 3 ] The larger K b, the stronger the base. - 11 -
Sample Problem 16.15 (p. 697) Calculate the concentration of OH - in a 0.15M solution of NH 3. (1.6 x 10-3 M) Practice Problem 1 (16.15) What is the ph of a 0.65 M solution of pyridine, C 5 H 5 N? K b = 1.7 x 10-9 a) 4.48 b) 8.96 c) 9.52 d) 9.62 e) 9.71 Practice Problem 2 (16.15) Which of the following compounds should produce the highest ph as a 0.05 M solution: pyridine, methylamine, or nitrous acid? (methylamine) - 12 -
Types of Weak Bases Weak bases generally fall into one of two categories. 1. Neutral substances with a lone pair of electrons that can accept protons. Most neutral weak bases contain nitrogen. Amines are related to ammonia and have one or more N H bonds replaced with N C bonds (e.g., CH 3 NH 2 is methylamine). Like NH 3, amines can abstract a proton from a water molecule by forming an additional N-H bond, as shown in this figure for methylamine. 2. Anions of weak acids are also weak bases. e.g.: ClO is the conjugate base of HClO (weak acid): ClO (aq) + H 2 O(l) HClO(aq) + OH (aq) K b = 3.3 x 10 7 Sample Exercise 16.16 (p. 698) A solution made by adding solid sodium hypochlorite (NaClO) to enough water to make 2.00 L of solution has a ph of 10.50. How many moles of NaClO were added to the water? (See info immediately above.) (0.60 mol) - 13 -
Practice Exercise 1 (16.16) The benzoate ion, C 6 H 5 COO -, is a weak base with K b = 1.6 x 10-10. How many moles of sodium benzoate are present in 0.50 L of a solution of NaC 6 H 5 COO if the ph is 9.04? a) 0.38 b) 0.66 c) 0.76 d) 1.5 e) 2.9 Practice Exercise 2 (16.16) What is the molarity of a solution of NH 3 in water has a ph of 11.17? (0.12 M) - 14 -
16.8 Relationship Between K a and K b Consider the following equilibria: NH 4 + (aq) NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l) NH 4 + (aq) + OH (aq) We can write equilibrium expressions for these reactions: K a [NH = 3 ][H [NH 4 ] [NH3 ] If we add these equations together: NH + 4 (aq) NH 3 (aq) + H + (aq) NH 3 (aq) + H 2 O(l) NH + 4 (aq) + OH (aq) The net reaction is the autoionization of water. H 2 O(l) H + (aq) + OH (aq) + + Recall that: K w = [H + ][OH ] K a x K b = K w Or pk a + pk b = pk w = 14.00 (at 25 o C) Thus, the larger K a (and the smaller pk a), the smaller K b (and the larger pk b ). The stronger the acid, the weaker its conjugate base and vice versa. ] K b [NH = + 4 ][OH - ] Sample Exercise 16.17 (p. 701) Calculate a) the base-dissociation constant, K b, for the fluoride ion (F - ); (1.5 x 10-11 ) b) the acid-dissociation constant, K a, for the ammonium ion (NH 4 + ). (5.6 x 10-10 ) - 15 -
Practice Exercise 1 (16.17) By using information from Appendix D, put the following three substances in order of weakest to strongest base: (i) (CH 3 ) 3 N (ii) HCOO - (iii) BrO - a) (i) < (ii) < (iii) b) (ii) < (i) < (iii) c) (iii) < (i) < (ii) d) (ii) < (iii) < (i) e) (iii) < (ii) < (i) Practice Exercise 2 (16.17) a) Which of the following anions has the largest base-dissociation constant: NO 2 -, PO 4 3-, or N 3 -? (PO 4 3-, K b = 2.4 x 10-2 ) b) The base quinoline has the following structure: Its conjugate acid is listed in handbooks as having a pk a of 4.90. What is the base-dissociation constant for quinoline? (7.9 x 10-10 ) - 16 -
16.9 Acid-Base Properties of Salt Solutions Acid-base properties of salts are a consequence of the reactions of their ions in solution. An Anion s Ability to React with Water Anions from weak acids are basic: They will cause an increase in ph. Anions from strong acids are neutral. They do not cause a change in ph. Anions with ionizable protons (e.g., HSO 4 ) are amphiprotic. They are capable of acting as an acid or a base. If K a > K b, the anion tends to decrease the ph. If K b > K a, the anion tends to increase the ph. A Cation s Ability to React with Water Polyatomic cations that have 1 or more ionizable protons are conjugate acids of weak bases decrease ph. Metal cations of Group 1A and heavy alkaline earth metals are cations of strong bases and do not alter ph. Other metal ions can cause a decrease in ph. Combined Effect of Cation and Anion in Solution The ph of a solution may be qualitatively predicted using the following guidelines: Salts derived from a strong acid and a strong base are neutral. Examples are NaCl and Ca(NO 3 ) 2. Salts derived from a strong base and a weak acid are basic. Examples are NaClO and Ba(C 2 H 3 O 2 ) 2. Salts derived from a weak base and a strong acid are acidic. An example is NH 4 Cl. Salts derived from a weak acid and a weak base can be either acidic or basic. Equilibrium rules apply! We need to compare K a and K b for hydrolysis of the anion and the cation. For example, consider NH 4 CN. Both ions undergo significant hydrolysis. Is the salt solution acidic or basic? The K a of NH 4 + is smaller than the K b of CN, so the solution should be basic. Sample Exercise 16.18 (p. 704) Determine whether aqueous solutions of each of the following salts will be acidic, basic, or neutral: a) Ba(CH 3 COO) 2, b) NH 4 Cl c) CH 3 NH 3 Br d) KNO 3 e) Al(ClO 4 ) 3-17 -
Practice Exercise 1 (16.18) Order the following solutions from lowest to highest ph: (i) 0.10 M NaClO (ii) 0.10 M KBr (iii) 0.10 M NH 4 ClO 4 a) (i) < (ii) < (iii) b) (ii) < (i) < (iii) c) (iii) < (i) < (ii) d) (ii) < (iii) < (i) e) (iii) < (ii) < (i) Practice Exercise 2 (16.18) In each of the following, indicate which salt in each of the following pairs will form the more acidic (or less basic) 0.010 M solution: a) NaNO 3 or Fe(NO 3 ) 3 b) KBr, or KBrO c) CH 3 NH 3 Cl or BaCl 2 d) NH 4 NO 2 or NH 4 NO 3 Sample Exercise 16.19 (p. 698) Predict whether the salt Na 2 HPO 4 will form an acidic or a basic solution on dissolving in water. (basic) - 18 -
Practice Exercise 1 (16.19) How many of the following salts are expected to produce acidic solutions (See Table 16.3 for data): NaHSO 4, NaHC 2 O 4, NaH 2 PO 4, NaHCO 3 a) 0 b) 1 c) 2 d) 3 e) 4 Practice Exercise 2 (16.19) Predict whether the dipotassium salt of citric acid (K 2 HC 6 H 5 O 7 ) will form an acidic or basic solution in water. (see Table 16.3 for data) (acidic) - 19 -
16.10 Acid-Base Behavior and Chemical Structure Factors That Affect Acid Strength The H X bond must be polar with H δ+ and X δ-. In ionic hydrides, the bond polarity is reversed. The H X bond is polar with H δ- and X δ+ : the substance is a base. Other factors important in determining acid strength include: The strength of the bond: the H X bond must be weak enough to be broken. The stability of the conjugate base, X : the greater the stability of the conjugate base, the more acidic the molecule. Binary Acids Group trends: The H X bond strength tends to decrease as the element X increases in size. Acid strength increases down a group; base strength decreases down a group. Period trends: Acid strength increases and base strength decreases from left to right across a period as the electronegativity of X increases. Oxyacids Acids that contain OH groups (and often additional oxygen atoms) bound to the central atom All oxyacids have the general structure Y O H, e.g. H 2 SO 4 The strength of the acid depends on Y and the atoms attached to Y. As the electronegativity of Y increases, so does the acidity of the substance. The bond polarity increases and the stability of the conjugate base (usually an anion) increases. Summary: For oxyacids with the same number of OH groups and the same number of oxygen atoms: Acid strength increases with increasing electronegativity of the central atom, Y. Example: HClO > HBrO > HIO For oxyacids with the same central atom, Y: Acid strength increases as the number of oxygen atoms attached to Y increases. Example: HClO 4 > HClO 3 > HClO 2 > HClO - 20 -
Strength of oxyacids Sample Exercise 16.20 (p.708) Arrange the compounds in each of the following series in order of increasing acid strength: a) AsH 3, HBr, KH, H 2 Se; b) H 2 SO 4, H 2 SeO 3, H 2 SeO 4. Practice Exercise 1 (16.20) Arrange the following substances in order from weakest to strongest acid: HClO 3, HOI, HBrO 2, HClO 2, HIO 2 a) HIO 2 < HOI < HClO 3 < HBrO 2 < HClO 2 b) HOI < HIO 2 < HBrO 2 < HClO 2 < HClO 3 c) HBrO 2 < HIO 2 < HClO 2 < HOI < HClO 3 d) HClO 3 < HClO 2 < HBrO 2 < HIO 2 < HOI e) HOI < HClO 2 < HBrO 2 < HIO 2 < HClO 3 Practice Exercise 2 (16.20) In each of the following pairs choose the compound that leads to the more acidic (or less basic) solution: a) HBr, HF; b) PH 3, H 2 S; c) HNO 2, HNO 3 ; d) H 2 SO 3, H 2 SeO 3. - 21 -
Carboxylic Acids There is a large class of acids that contain a COOH group (a carboxyl group). Acids that contain this group are called carboxylic acids: e.g. acetic acid, benzoic acid, formic acid. These molecules are acidic because: 1. The additional oxygen atom on the carboxyl group increases the polarity of the O H bond and stabilizes the conjugate base. 2. The conjugate base exhibits resonance. This gives it the ability to delocalize the negative charge over the carboxylate group, further increasing the stability of the conjugate base. The acid strength also increases as the number of electronegative groups in the acid increases. For example, acetic acid is much weaker than trichloroacetic acid. Comparison of Different Types of Acids and Bases Arrhenius (traditional) acids and bases (C19th) Acid: compound containing H that ionizes to yield H + in solution Base: compound containing OH that ionizes to yield OH - in solution (Note: does not describe acid/base behavior in solvents other than water) Note: Every Arrhenius acid/base is also a Brønsted-Lowry acid/base. Brønsted-Lowry Acids and Bases (1923) Acid: H + (proton) donor Base: H + (proton) acceptor NH 3 + H 2 O + NH 4 + OHammonia water ammonium ion hydroxide ion (B-L base) (B-L acid) (B-L acid) (B-L base) Note: Every Brønsted-Lowry acid/base is also a Lewis acid/base. Lewis Acids and Bases (1920 s) Acid: accepts pair of e- s Base: donates pair of e- s H + + [:O:H] - H:O:H Lewis acid Lewis base - 22 -
16.11 Lewis Acids and Bases A Brønsted-Lowry acid is a proton donor. Lewis: emphasize the shared electron pair. A Lewis acid is an electron pair acceptor. A Lewis base is an 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. What types of compounds can act as Lewis acids? Lewis acids generally have an incomplete octet (e.g., BF 3 ). For a substance to be a proton acceptor (a Brønsted- Lowry base), it must have an unshared pair of electrons for binding the proton. NH 3, for example, acts as a proton acceptor. Using Lewis structures, we can write the reaction between H + and NH 3 as shown in the firugre. Consider the reaction between NH 3 and BF 3. This reaction occurs because BF 3 has a vacant orbital in its valence shell. It therefore acts as an electron-pair acceptor (a Lewis acid) toward NH 3, which donates the electron pair. The curved arrow shows the donation of a pair of electrons from N to B to form a covalent bond. Transition-metal ions are generally Lewis acids. Lewis acids must have a vacant orbital (into which the electron pairs can be donated). Compounds with multiple bonds can act as Lewis acids. For example, consider the reaction: H 2 O(l) + CO 2 (g) H 2 CO 3 (aq) Water acts as the electron pair donor and carbon dioxide as the electron pair acceptor in this reaction. Overall, the water (Lewis base) has donated a pair of electrons to the CO 2 (Lewis acid). - 23 -
Hydrolysis of Metal Ions The Lewis concept may be used to explain the acid properties of many metal ions. Metal ions are positively charged and attract water molecules (via the lone pairs on the oxygen atom of water). Hydrated metal ions act as acids, e.g. Fe(H 2 O) 3+ 6 (aq) Fe(H 2 O) 5 (OH) 2+ (aq) + H + (aq) K a = 2 x 10 3. In general: The higher the charge, the stronger the M OH interaction: 2 K a values generally increase with increasing charge The smaller the metal ion, the more acidic the ion. K a values generally decrease with decreasing ionic radius Thus the ph of an aqueous solution increases as the size of the ion increases (e.g., Ca 2+ vs. Zn 2+ ) and as the charge increases (e.g., Na + vs. Ca 2+ and Zn 2+ vs. Al 3+ ). The Amphoteric Behavior of Amino Acids Amino acids: building blocks of proteins. Each contains a carboxyl group AND an amine group. Thus amino acids have both acidic and basic groups. They undergo a proton transfer in which the proton of the carboxyl is transferred to the basic nitrogen atom of the amine group A zwitterion or dipolar ion results. - 24 -