CHEM1112 Lecture Notes - Lecture 7: Conjugate Acid, Endothermic Process, Equilibrium Constant

Chem 1030 Chapter 16 Chemistry of Acids and Bases Exercise 1 Name: 1. What is the pH of a solution which has an (OH ] concentration of 1.3 x I 0-M? Is this more or less basic than pure water? 2. Identify each of the compounds below as an acid or a base. For the acids, identify the conjugate base of the acid. For the bases, identify the conjugate acid of the base. COMPOUND H2SO4 OH HS- H S HPO,2 ACID OR BASE? CONJUGATE... 3. Complete the reaction that occurs when each of the following is placed in water i) identify the acid and the base on gach side of the equation; ii) identify the conjugate acid-base pairs in each equation. i) Write the equilibrium constant expression K or K, for each.

Adding a strong Acid to a Buffer Part A Learning Goal : A beaker with 145 mL of an acetic acid buffer with a pH of 5,000 is sitting on a benchtop. The total molarity of acid and conjugate base in this buffer is 0.100 M. A student adds 8.60 mL of a 0.320 M HC1 solution to the beakerHow much will the pH change? The pK, of acetic acid is 4.740. To understand how buffers use reserves of conjugate acid and conjugate base to counteract the effects of acid or base addition on pH Express your answer numerically to two decimal places. Use a minus (-) sign if the pH has decreased. Hints A buffer is a mixture of a conjugate acid-base pair. In other words, it is a solution that contains a weak acid and its conjugate base, or a weak base and its conjugate acid. For example, an acetic acid buffer consists of acetic acid, CH, COOH, and its conjugate base, the acetate ion CH3 COO. Because ions cannot simply be added to a solution, the conjugate base is added in a salt form (eg., sodium acetate NaCH COO). IVO Aze s o o o ? ApH -- 0.363 Submit My Answers Give Up Incorrect, Try Again; 5 attempts remaining Buffers work because the conjugate acid-base pair work together to neutralize the addition of H' or OH ions. Thus, for example, if H ions are added to the acetate buffer described above, they will be largely removed from solution by the reaction of Hwith the conjugate base: Provide Feedback Continue H+ + CH, COO CH, COOH Similarly, any added OH ions will be neutralized by a reaction with the conjugate acid OH + CH3 COOHCHCOO + H2O This buffer system is described by the Henderson-Hasselbalch equation [conjugate base] pH = pK. log [conjugate acid)

Ka values for strong acids and bases are so large that you will often see the reaction written with a single arrow ("completely dissociated in solution") as if there was no back reaction. I would prefer for you to think of it as an equilibrium with such a large equilibrium constant that the back reaction is negligible. When we want to write a Ka or a Kb expression, we react the acid or base directly with water and write out the equilibrium. To quantify the strength of the conjugate base C in aqueous solution, we look at the reaction of just the this ion in water by itself. The CI will be on the reactant side and is now defined as the base in the reaction. The base, Cl (proton acceptor) reacts with water by accepting the Ht ion to form the new products. Could you write this equation? Take a minute to write out this equation below Enter your answer in the order: base + acid yields conjugate acid + conjugate base. To save time, you can skip the phase labels this time. Since you are working with a base (CI), the equilibrium constant is now Kb. chemPad ⓤ Help [Ci] Greek ▼ Look at what happens when we multiply the Ka for the acid times the Ko for its conjugate base. Everything cancels except the terms for [O] and [H3O+]. The product of [OH-] X [H30*) is the water constant or Kw (1.0 x 10.1 Ka X Kb [HCI] [ci] = [OH-][H3O+] = Kw = 1.0 X 10-14 Using the water constant, we can calculate the K, for any conjugate base if we know the Ka for the acid (and vice versa).