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eight equations, you will always be given the Kaor Kb. Some K's that are worth memorizing are Kafor acetic acid = 1.8 x 10-5, and Kbfor NH3= 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-+ H2O ←→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 H3O+, then we could not call it Kcb). Let's call it Kcb. There is an equation for calculating Kcb. Kw= Kax Kcb. Therefore, Kcb= (Kw/Ka). Kwis the hydrolysis constant for water at 25oC. It is 1 x 10-14. You will always be given Ka(or Kb) if you need that for a calculation. Kcb= (1 x 10-14)/(1.8 x 10-5) = 5.56 x 10-10.Let's look at the second equation:A-+ H3O+→HA + H2OThis is a reaction of a conjugate base with a strong acid. (Note that each of the seven strong acids react with H2O to form H3O+, so H3O+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 Kcbsa(Kconjugatebasestrongacid). 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, Kcbsa= 1/Ka= 1/(1.8 x 10-5) = 5.6 x 104. 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-→H2O + A-This is a reaction of a weak acid and a strong base. Let's call the K for this reaction Kwasb= Kweakacidstrongbase. 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 = Kwasb= 1/Kcb= 1/(Kw/Ka) = Ka/Kw= (1.8x10-5)/(1x10-14) = 1.8 x 109. Note that a reaction with a strong base has a large K.
Chem 162-2016 Chapter 17 Tro Acid-Base Equilibria 54SEVEN STRONG ACIDStransparencyET: 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 KaAcid spectator ion)* ~106HCl + H2O H3O++ Cl-Kcb= ~10-23 ~108HBr + H2O H3O++ Br-Kcb= ~10-22~109HI + H2O H3O++ I-Kcb= ~10-20~103H2SO4+ H2O H3O++ HSO4-**Kcb= ~10-17~2x101HNO3+ H2O H3O++ NO3-Kcb= ~10-15~1010HClO4+ H2O H3O++ ClO4-Kcb= ~10-24~101HClO3+ H2O H3O++ ClO3-Kcb= ~10-15*These conjugate bases are known as “nominal bases”, i.e., a base by definition, but effectively not a base. **Although HSO4-is effectively not a base, it iseffective an acid. HSO4-+ H2O ←→H3O++ SO42-All other acids are weak acids, e.g., HA (acetic acid), H2SO3, RNH3+, BF3CH3COOH + H2O H3O++ CH3COO-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