eight equations you will always be given the K a or K b Some Ks that are worth

Eight equations you will always be given the k a or k

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eight equations, you will always be given the K a or K b . Some K's that are worth memorizing are K a for acetic acid = 1.8 x 10 -5 , and K b for NH 3 = 1.8 x 10 -5 . The fact that they are the same is a coincidence. You will never be given the K's of the other six equations. You'll need to calculate them. Let's look at the third equation: A - + H 2 O HA + OH - This is a reaction of a conjugate base with water (notice that it is uncharged water; if it was charged water, such as H 3 O + , then we could not call it K cb ). Let's call it K cb . There is an equation for calculating K cb . K w = K a x K cb . Therefore, K cb = (K w /K a ). K w is the hydrolysis constant for water at 25 o C. It is 1 x 10 -14 . You will always be given K a (or K b ) if you need that for a calculation. K cb = (1 x 10 -14 )/(1.8 x 10 -5 ) = 5.56 x 10 -10 . Let's look at the second equation: A - + H 3 O + HA + H 2 O This is a reaction of a conjugate base with a strong acid. (Note that each of the seven strong acids react with H 2 O to form H 3 O + , so H 3 O + becomes the new strong acid in water.) There is no name for the K for this reaction. Let's give it a name. Let's call it K cbsa (K conjugatebasestrongacid ). What is the equilibrium constant for this reaction? In looking at the right side of the reaction note that the right side of the equation is a weak acid + water. Hence, this reaction is the reverse of the first reaction, weak acid + water. Since the reaction is reversed, then the K must be inverted. Hence, K cbsa = 1/K a = 1/(1.8 x 10 -5 ) = 5.6 x 10 4 . Note that reactions with strong acids (or strong bases) have large K's, and reactions with weak acids (or weak bases) have small K's. Let's look at the fourth equation: HA + OH - H 2 O + A - This is a reaction of a weak acid and a strong base. Let's call the K for this reaction K wasb = K weakacidstrongbase . What is the K for this reaction? Look on the right side of this reaction. Looks familiar? That's the reaction of a conjugate base with water. Hence, this equation is the reverse of the third equation. Since the third equation is reversed, then the K needs to be inverted. So the K for this equation = K wasb = 1/K cb = 1/(K w /K a ) = K a /K w = (1.8x10 -5 )/(1x10 -14 ) = 1.8 x 10 9 . Note that a reaction with a strong base has a large K.
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Chem 162-2016 Chapter 17 Tro Acid-Base Equilibria 54 SEVEN STRONG ACIDS transparency ET: Discuss strong acids and strong bases and the strengths of their conjugate bases and acids; practice both. Very weak conj. base (i.e., not a base but a K a Acid spectator ion)* ~10 6 HCl + H 2 O H 3 O + + Cl - K cb = ~10 -23 ~10 8 HBr + H 2 O H 3 O + + Br - K cb = ~10 -22 ~10 9 HI + H 2 O H 3 O + + I - K cb = ~10 -20 ~10 3 H 2 SO 4 + H 2 O H 3 O + + HSO 4 -** K cb = ~10 -17 ~2x10 1 HNO 3 + H 2 O H 3 O + + NO 3 - K cb = ~10 -15 ~10 10 HClO 4 + H 2 O H 3 O + + ClO 4 - K cb = ~10 -24 ~10 1 HClO 3 + H 2 O H 3 O + + ClO 3 - K cb = ~10 -15 *These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base. **Although HSO 4 - is effectively not a base, it is effective an acid. HSO 4 - + H 2 O H 3 O + + SO 4 2- All other acids are weak acids, e.g., HA (acetic acid), H 2 SO 3 , RNH 3 + , BF 3 CH 3 COOH + H 2 O H 3 O + + CH 3 COO - Note inverse relationship between the strength of the acid and the strength of the conjugate base. A strong acid (e.g., HCl) has a very weak conjugate base (Cl
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