21.1 5.111 Lecture Summary #21 AcidBase Equilibrium Read Chapter 10 Topics: Classification of AcidBases, Autoionization of Water, p unction, Strength of Acids and Bases, Equilibrium Involving Weak Acids. Classification of Acids and Bases 1. Arrhenius a narrow definition of acids and bases An acid is a substance that when dissolved in water increases the concentration of hydrogen ions. A base is a substance that increases the hydroxide concentration. 2. BrØnstedLowry a broader definition A BrØnstedLowry acid a substance that can donate a hydrogen ion A BrØnstedLowry base a substance that can accept a hydrogen ion Example 1 C 3 COO (aq) 2 O (l) 3 O (aq) C 3 COO (aq) Acid1 Base2 Acid2 Base1 (note: hydronium ion 3 O (aq) is used instead of (aq) to represent the true nature of hydrogen ions in water) Acidbases occur as conjugate acidbase pairs. C 3 COO and C 3 COO are a pair. 2 O and 3 O are a pair. The conjugate base of an acid is the base that is formed when the acid has donated a hydrogen ion. The conjugate acid of a base is the acid that forms when base accepts a hydrogen ion. Example 2 Which are BrØnstedLowry acids and which are BrØnstedLowry bases? CO 3 (aq) 2 O (l) 3 O (aq) CO 3 2 (aq) CO 3 (aq) 2 O (l) 2 CO 3 (aq) O (aq) amphoteric molecules that can function either as acids or bases depending on the reaction conditions.
21.2 3. Lewis Acid and Base more general definition applies to reactions that don't involve a hydrogen ion Lewis base species that donates lonepair electrons Lewis acid species that accepts such electrons Example 1 B N B N Ammonia is the Lewis base. It donates lonepair electrons to B 3, the Lewis acid and the electron acceptor. Autoionization of Water 2 O (l) 2 O (l) 3 O (aq) O (aq) or 2 2 O (l) 3 O (aq) O (aq) acid base acid base ow much 2 O is in a glass of water? G = G f ( 3 O,aq) G f (O,aq) 2 G f ( 2 O,l) = (237.13) (157.24) 2 x (237.13) kj/mol = 79.89 kj/mol ln K =!G /RT = ( 7.989 x 10 4 J/mol) = 32.24 (8.3145 J/Kmol)(298.0 K) K = 1.0 x 10 14 at 298 K This very small value indicates that only a small proportion of water molecules are ionized. Concentration of ions due to autoionization of water is very low, about 1 molecule in 200 million.
21.3 K = [ 3 O ][O ] This K is called K w. Because K w is an equilibrium constant, the product of [ 3 O ][O ] is always 1.0 x 10 14 at 298 K. Note: Because the concentration of the solvent, 2 O, does not change significantly in a dilute solution, it does not enter the equilibrium expression. The solvent, water, is very nearly pure, and pure liquids and pure solids are not included in equilibrium expressions. p unction p = log [ 3 O ] po unction po = log [O ] K w = [ 3 O ][O ] logk w = log[ 3 O ] log[o ] logk w = log[ 3 O ] log[o ] pk w = p po = 14.00 at 25 C Strength of Acids and Bases p of pure water p = log (1.0 x 10 7 ) = 7.00 p of an acid solution is p of an base solution is EPA defines waste as "corrosive" if the p is lower than 3.0 or higher than 12.5. 1. Acid in water C 3 COO (aq) 2 O (l) 3 O (aq) C 3 CO 2 (aq) Acid ionization constant K a = [ 3 O ] [C 3 CO 2 ] [C 3 COO] K a equals 1.76 x 10 5 at 25 C. Small value tells us that only a small proportion of C 3 COO molecules donate their proton when dissolved in water (weak acid).
21.4 A (aq) 2 O (l) 3 O (aq) A (aq) ACID (A) IN WATER B (aq) 2 O (l) 3 O (aq) B (aq) ACID (B ) IN WATER A strong acid has a K a >1 which means that the acid ionizes almost completely. A weak acid has a K a <1. The reaction with water does not produce many ionized species before equilibrium is reached. pk a = log K a The lower the value of K a, the higher the value of pk a. The higher the pk a, the weaker the acid. 2. Base in water N 3 (aq) 2 O (l) N 4 (aq) O (aq) Base ionization constant K b = [N 4 ] [O ] [N 3 ] K b is 1.8 x 10 5 at 25 C. This small value tells us that only a small amount of N 3 ionizes to N 4 and O in solution. A strong base reacts essentially completely to give O (aq) when put in water. N 3 is not a strong base. It is a moderately weak base. B (aq) 2 O (l) B (aq) O (aq) BASE (B) IN WATER A (aq) 2 O (l) A (aq) O (aq) BASE (A ) IN WATER pk b = log K b larger K b, stronger base larger pk b, weaker base 3. Conjugate acids and bases The stronger the acid, the weaker its conjugate base. The stronger the base, the weaker its conjugate acid. Consider conjugate acidbase pair N 3 and N 4 : N 3 (aq) 2 O (l) N 4 (aq) O (aq) N 4 (aq) 2 O (l) 3 O (aq) N 3 (aq) Multiply K's together and get: K a x K b = [N 3 ][ 3 O ] x [N 4 ][O ] [N 4 ] [N 3 ] = [ 3 O ][O ]
21.5 K a x K b = K w log K a log K b = log K w or pk a pk b = pk w = 14.00 Strong acid A (aq) 2 O (l) Strong base B (aq) 2 O (l) 3 O (aq) A (aq) B (aq) O (aq) 4. Relative strengths of acids Is NO 3 or N 4 a stronger acid? Will the reaction lie far to the right or left? NO 3 (aq) N 3 (aq) NO 3 (aq) N 4 (aq) [NO 3 ][N 4 ] K = [NO 3 ][N 3 ] consider each acid separately: 1. NO 3 (aq) 2 O (l) 3 O (aq) NO 3 (aq) [ 3 O ][NO 3 ] K a(no 3 ) = = 20. [NO 3 ] 2. N 4 (aq) 2 O (l) 3 O (aq) N 3 (aq) [ 3 O ][N 3 ] K a (N 4 ) = = 5.6 x 10 10 [N 4 ] Subtract equation 2 from 1 and divide the corresponding equilibrium constants. [ 3 O ][NO 3 ] K a (NO 3 ) [NO 3 ] [NO 3 ][N 4 ] K = = 20. = = K a (N 4 ) [ 3 O ][N 3 ] [NO 3 ][N 3 ] 5.6 x 10 10 [N 4 ] = 3.6 x 10 10 Reaction lies far to the. NO 3 is a than N 4.
21.6 Types of acidbase problems 1. weak acid in water 2. weak base in water salt in water 3. strong acid in water 4. strong base in water 5. buffer Equilibrium involving weak acids Example: Vitamin C (ascorbic acid, C 6 7 O 6 ) has a K a of 8.0 x 10 5. Calculate the p of a solution made by dissolving 500. mg in 100. ml of water. 0.500 g x 1 mol/176.126 g = 2.84 x 10 3 mol 2.84 x 10 3 mol/0.100 L = 0.0284 M C 6 7 O 6 (aq) 2 O (l) 3 O (aq) C 6 7 O 6 (aq) C 6 7 O 6 3 O C 6 7 O 6 initial molarity 0.0284 0 0 change in molarity x x x equilibrium molarity 0.0284 x x x K a = 8.0 x 10 5 = [ 3 O ][C 6 7 O 6 ] [C 6 7 O 6 ] = x 2 0.0284 x If x<< 0.0284, then (0.0284x) ~= 0.0284. K = 8.0 x 10 5 = a x 2 0.0284 x = 0.00151 (really 2 sf, but carry extra) Check assumption. Is 0.0284 0.00151 ~= 0.0284? You can use assumption if x is less than 5% of the value in question. ere (0.00151/0.0284) x 100% = 5.3% (more than 5%), so must use the quadratic equation. Using quadratic eq, x = 0.00147 (really 2 sf) p = log [1.47 x 10 3 ] = 2.83
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