Chpt 16 Acids and Bases
Defining Acids Arrhenius: Acid: Substances when dissolved in water increase the concentration of H+. Base: Substances when dissolved in water increase the concentration of OH- Brønsted-Lowry: Acid: substance that donates a proton to another substance Base: substance that accepts a proton from another substance. Lewis: Acid: substance that can accept an electron pair Base: substance that can donate an electron pair Amphiprotic: substance that can be both acid and base
Conjugate acid-base pairs Substance can function as an acid only if there is a substance to act as a base. Remove H + HCl (aq) + H 2 O (l) Cl - (aq) + H 3 O + (aq) Add H + An acid and base that differ by the presence or absence of a proton are a conjugate acid-base pair.
Strong Acids and Bases The stronger an acid, the weaker its conjugate base. The stronger a base, the weaker its conjugate acid For the reaction NH 3 (aq) + H 2 O( ) NH 4+ (aq) + OH - (aq) NH 4 + and H 2 O both act as Brønsted-Lowry acids (they donate a proton). OH - and NH 3 behave as bases (proton acceptors) OH - is a strong base, H 2 O is a weak acid. NH 3 is a weaker base than OH -, therefore NH 4 + is a slightly stronger acid than H 2 O The position of the equilibrium will favor transfer of a proton from the stronger acid to the stronger base the above equilibrium will lie to the left
Strong Acids and Bases A strong acid is a substance which completely dissociates into H + and X - ions in solution. HNO 3 (aq) H + (aq) + NO 3- (aq) The most common strong acids include: HNO 3, HBr, HI, HCl, H 2 SO 4 A strong base is a substance which completely dissociates (or reacts) to form OH - and X + (X) NaOH(aq) Na + (aq) + OH - (aq) N 3- (aq) + 3H 2 O( ) NH 3 (aq) + 3OH - (aq) The most common strong bases include: NaOH, KOH, Ca(OH) 2, O 2-, H -, N 3- The [H + ] or [OH - ] can be determined from the initial strong acid or base concentration
Autoionization of Water The dissociation of water can be represented by the equilibrium H 2 O( ) H + (aq) + OH - (aq) The equilibrium expression is K w = [H + ][OH - ] K w is the ion-product constant and at 25 o C is 1.0*10-14 Assumed to be valid for any dilute solution so one can calculate the concentration of H + and OH - if the other is known.
ph scale For a neutral solution the [H + ] = [OH - ]. [H + ] 2 = 1.0*10-14 [H + ] = 1.0*10-7 M More convenient to state [H + ] on a log scale, called the ph scale ph=-log[h+] ph of a neutral solution is 7.00. Type [H+] [OH-] ph poh Acidic >1.0x10-7 <1.0x10-7 < 7.00 > 7.00 Neutral =1.0x10-7 =1.0x10-7 = 7.00 = 7.00 Basic <1.0x10-7 >1.0x10-7 > 7.00 < 7.00
ph A neutral solution of pure water with a ph of 7 is heated to 50 o C. The ph drops to 6.63. (T/F) The concentration of [H + ] is greater than the concentration of [OH - ]. A True B False
Strong Acids and Bases Example: What is the ph of a 0.040 M solution of HCl? Example: How many moles of KOH is required to create a 1-L solution with a ph of 11.89? A 1.3 x 10-12 B 7.7 x 10-3 C 3.9 x 10-7 HCl(aq) H+(aq) Cl-(aq) Initial 0.040 0 0 Change -0.040 +0.040 +0.040 Equilibrium 0 +0.040 +0.040 ph = -log [H + ] ph = -log (0.040) = 1.40 D not enough information
Weak Acids Most acidic substances are weak acids and only partially dissociate in aqueous solution HA(aq) H + (aq) + A - (aq) K a K a is the acid-dissociation constant Acid Formula K a hydrofluoric HF 6.8 x 10-4 nitrous HNO 2 4.5 x 10-4 acetic HC 2 H 3 O 2 1.8 x 10-5 phenol HOC 6 H 5 1.3 x 10-10 [H ][A ] [HA] -
Calculating K a Example: A 0.020 M solution of niacin (one of the B-vitamins) has a ph of 3.26. Calculate K a for this substance. [H + ] = 10 -ph = 10-3.26 = 5.49 x 10-4 M [C 6 H 4 NO 2- ] = [H + ] = 5.49 x 10-4 M HC 6 H 4 NO 2 (aq) H + (aq) C 6 H 4 NO 2- (aq) Initial 0.020 0 0 Change 0.020-5.49*10-4 5.49*10-4 5.49*10-4 Final ~0.020 5.49*10-4 5.49*10-4 K a -4 [5.49x10 ][5.49x10 [0.020] -4 ] K a = 1.51 x 10-5 percent ionization = [H + ]*100/[HC 6 H 4 NO 2 ] = (5.49x10-4 )/(0.020) = 2.75%
Calculating ph of a Weak Acid I Calculate the ph of a 0.010 M HF solution (K a ~ 1*10-3 ). HF(aq) H + (aq) F - (aq) Initial 0.010 0 0 Change -x +x +x Final 0.01-x x x Is our original assumption valid? Ka = x 2 /(0.01-x) Assume x is small Ka = x 2 /0.01 X = 0.003 X is not small [H+] = 0.003 ph = -log[0.003] = 2.5
Calculating ph of a Weak Acid II Calculate the ph of a 0.010 M HF solution (K a ~ 1*10-3 ). HC 6 H 4 NO 2 (aq) H + (aq) C 6 H 4 NO 2- (aq) Initial 0.010 0 0 Change -x +x +x Final 0.01-x x x Ka = x 2 /(0.01-x) Ka(x 2-0.02x+0.0001) =x 2 0.999x 2 + 2e -5 x -1e -7 = 0.0 X = (-b +- sqrt(b2 4ac))/2a X = (-2e-5 +- sqrt(4e-10 + 3.996e-7)/1.998 X = 3.06e-4, -3.26e-4 3.06e-4/0.01 = 0.03 = 3% error X = [H] = 3.06e-4 ph = -log[3.06e-4] =3.5
Calculating ph of a Weak Acid III Example: Calculate the ph of a 0.010 M solution of niacin, K a = 1.51 x 10-5. A 3.41 B 4.82 C 6.82 D 9.13 E not enough information
Polyprotic acids Acids which can donate more than one proton are known as polyprotic acids each hydrogen atom is ionized in successive steps I. H 3 PO 4 (aq) H 2 PO 4- (aq) + H + (aq) II. H 2 PO 4- (aq) HPO 4 2- (aq) + H + (aq) III. HPO 4 2- (aq) PO 4 3- (aq) + H + (aq) Each step has its own value of K a I. K a = 7.5 x 10-3 II. K a = 6.2 x 10-8 III. K a = 4.2 x 10-13 It is successively more difficult to remove protons from a polyprotic acid K a decreases with each step
ph of polyprotic acids I Example: Determine the ph of a 0.020 M phosphoric acid (H 3 PO 4 ) solution. Since K a1 is so much larger than the other aciddissociation constants, the [H + ] will originate mainly from the first dissociation reaction. H 3 PO 4 (aq) H 2 PO 4- (aq) H + (aq) Initial 0.020 0 0 Change -x +x +x Final 0.020 - x +x +x
ph of polyprotic acids II H 3 PO 4 (aq) H 2 PO 4- (aq) H + (aq) Initial 0.020 0 0 Change -x +x +x Final 0.020 - x +x +x -3 7.5 x 10 K a 1 [x][x] [0.020] x 2 = 7.5 x 10-3 (0.020) x = 0.0122 M % dissociation = (0.0122/0.02)x100 > 50% our assumption that x is small invalid!
ph of polyprotic acids IV H 3 PO 4 (aq) H 2 PO 4- (aq) H + (aq) Initial 0.020 0 0 Change -0.00905 0.00905 0.00905 Final 0.01095 0.00905 0.00905 [H + ] = 9.05 x 10-3 ph = - log [H + ] ph = -log(9.05 x 10-3 ) ph = 2.04 What is the proton concentration from the second dissociation reaction? H 2 PO 4- (aq) HPO 4 2- (aq) + H + (aq) K a2 = 6.2 x 10-8
You find an unusual acid of the form HAB and are investigating its properties for possible patents. You create a 0.01 M solution of the acid in water and the principle equilibrium established is HAB H + (aq) + A (aq) + B - (aq) Calculate the ph of the solution. K a = 1 x 10-10 A 2 B 4 C 6 D 8 E not enough information
Question Carbonic acid H 2 CO 3 plays a major role in the circulatory system. At the ph of blood, 7.4, the concentration of [H 2 CO 3 ] and [HCO 3- ] were measured to be 0.0231 M and 0.25 M respectively. What is the value of K a1? A 4 x 10-11 B 4 x 10-9 C 4 x 10-7 D not enough information
Weak Bases Weak bases partially dissociate in solution to form the appropriate conjugate acid and OH - ions A - + H 2 O( ) HA(aq) + OH - (aq) The equilibrium expression is given as K b [HA][OH - [A ] K b is the base-dissociation constant Base Formula K b ammonia NH 3 1.8 x 10-5 pyridine C 5 H 5 N 1.7 x 10-9 fluorine ion F - 1.5 x 10-11 - ]
K b of a weak base Calculate K b of codeine (C 18 H 21 NO 3 ) if a 0.005 M solution has a poh of 4.05. C 18 H 21 NO 3 (aq) C 18 H 22 NO 3+ (aq) OH - (aq) Initial 0.005 0 0 Change Final
ph of a weak base I Example: Calculate the ph of a 0.010 M solution of ammonia. K b = 1.8*10-5 NH 3 (aq) NH 4+ (aq) OH - (aq) Initial 0.010 0 0 Change -x +x +x Final 0.010 - x +x +x A 3.4 B 6.8 C -5 [x][x] K b 10.6 1.8 x 10 D 14.0 E not enough information [0.010] x 2 = 1.8 x 10-5 (0.010) x 2 = 1.8 x 10-7 x = 4.2 x 10-4
K a and K b Qualitatively, we have discussed the relationship between an acid and its conjugate base the stronger an acid, the weaker its conjugate base Consider the following reactions HF(aq) H + (aq) + F - (aq) K a = 6.8 x 10-4 F - (aq) + H 2 O( ) HF(aq) + OH - (aq) K b = 1.5 x 10-11 The overall reaction for the above is H 2 O( ) H + (aq) + OH - (aq) The overall equilibrium constant is the product of the individual reaction constants K = K a K b K w = (6.8 x 10-4 )(1.5 x 10-11 ) = 1.0 x 10-14
Salt Solutions Dissolved salt solutions can be acidic or basic. Ions can react with water to generate H + (aq) or OH - (aq) - hydrolysis To determine the extent to which an ion will react with water inspect the conjugate acid/base. Cl - : conjugate acid HCl is a strong acid, Cl - will not affect ph of solution F - : conjugate acid HF is a weak acid, F - will create a slighlty basic solution NH 4+ : conjugate base is NH 3 is a weak base, NH 4 will create a slightly acidic solution. The behavior of amphiprotic ions is determined by relative magnitude of K a and K b
ph of Salt Solutions Example: Calculate K a for Al(NO 3 ) 3 if a 1.0 M solution of this substance has a ph = 3.5 The Al 3+ metal ion will hydrolyze water and create H + ions. NO 3 - is the salt of a strong acid and is neutral in solution Al 3+ + H 2 O( ) AlOH 2+ (aq) + H + (aq) K a [H ][AlOH [Al 3 ] 2 ] [H+] = 10 -ph = 10-3.5 = 3.16 x 10-4 M [AlOH 2+ ] = [H + ] = 3.16 x 10-4 M [Al 3+ ] = 1.0 M [3.16e - 4][3.16e - 4] [1.0] K K a = 1.0 x 10-7 a
Question A solution of NaF is formed. Which of the following pictures best best represents aqueous NaF? The water molecules are omitted for clarity. Na + F - OH - HF A B C
Structure and acid behavior X H The bond must be polarized with a positive charge on the H The weaker the bond the stronger the acid The more stable the conjugate base the stronger the acid Oxyacids If central atom is a metal, larger electronegativity differences lead to stronger bases If central atom is a non-metal, as electronegativity increases the acidity increases. Carboxylic acids Acids due to presence of additional oxygen