Concept explainers
Interpretation: In the given molecular model, an aqueous solution of two conjugate acid-base pairs
Concept introduction: Strong acid can donate its proton. Strong acid/base will have weaker conjugate base/acid. The equilibrium constant
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CHEMISTRY: THE MOLECULAR NATURE OF MATTE
- For conjugate acidbase pairs, how are Ka and Kb related? Consider the reaction of acetic acid in water CH3CO2H(aq)+H2O(l)CH3CO2(aq)+H3O+(aq) where Ka = 1.8 105 a. Which two bases are competing for the proton? b. Which is the stronger base? c. In light of your answer to part b. why do we classify the acetate ion (CH3CO2) as a weak base? Use an appropriate reaction to justify your answer. In general, as base strength increases, conjugate acid strength decreases. Explain why the conjugate acid of the weak base NH3 is a weak acid. To summarize, the conjugate base of a weak acid is a weak base and the conjugate acid of a weak base is a weak acid (weak gives you weak). Assuming Ka for a monoprotic strong acid is 1 106, calculate Kb for the conjugate base of this strong acid. Why do conjugate bases of strong acids have no basic properties in water? List the conjugate bases of the six common strong acids. To tie it all together, some instructors have students think of Li+, K+, Rb+, Cs+, Ca2+, Sr2+, and Ba2+ as the conjugate acids of the strong bases LiOH, KOH. RbOH, CsOH, Ca(OH)2, Sr(OH)2, and Ba(OH)2. Although not technically correct, the conjugate acid strength of these cations is similar to the conjugate base strength of the strong acids. That is, these cations have no acidic properties in water; similarly, the conjugate bases of strong acids have no basic properties (strong gives you worthless). Fill in the blanks with the correct response. The conjugate base of a weak acid is a_____base. The conjugate acid of a weak base is a_____acid. The conjugate base of a strong acid is a_____base. The conjugate acid of a strong base is a_____ acid. (Hint: Weak gives you weak and strong gives you worthless.)arrow_forwardDetermine the equilibrium constant for the following reaction: Ca(OH)2(s) + 2 H+(aq) --> Ca2+(aq) + 2 H2O(l) given the chemical reactions below. Ca(OH)2(s) --> Ca2+(aq) + 2 OH-(aq) K = 6.5 X 10-6 H2O(l) --> H+(aq) + OH-(aq) K = 1.0 X 10-14arrow_forwardHA, a weak acid, also forms the ion HA, in solution. The reaction is HA(aq) + A (aq) = HA, (aq) The weak acid has K, = 8.64 × 10-7 and the reaction has K = 3.80. Calculate the concentrations of A, H*, and HA, in a 1.00 M solution of HA. [A¯] = M [H*] = M [HA,] = Marrow_forward
- Hydrazoic acid, HN3, has an acid dissociation constant of 2.5 x 10-5. Calculate the equilibrium concentrations of all substances if the initial concentration of HN3 is 0.0750 M. Determine the pH of the solution. Would a 0.0750 M solution of HBr have a higher or lower pH than the 0.0750 M HN3 solution? Explain why.arrow_forward9. We prepared a solution containing 1.0 mol.L' HF and 1.0 mol.L·' C,H3OH. The K, of hydrofluoric acid and phenol are 7.2 x 104 and 1.6 x 10-10 respectively. What is the pH of this solution? (Hint: find which acid is strongest and make the appropriate assumptions) b. What is the concentration of C,H;O¯ at equilibrium? а.arrow_forwardWhat is one acidifying process that produces H+ ions?arrow_forward
- What is the equilibrium expression for the following reaction? Zn(s) + 2NAOH(aq) + 2H2O(e) = NazZn(OH)4(aq) + H2(g)arrow_forwardGiven 0.01 M solutions of each of the following acids, which solution would have the lowest pH? -117 Hypoiodous acid (HOI), K = 2.3 x 10 Hypobromous acid (HOBr), K = 2.5 x 10 Lactic acid (HC₂H₂O₂), K = 1.3 x 10 Chlorous acid (HClO₂), K = 1.1 x 10²arrow_forwardWhich of the following would decrease the concentration of Pb+2 ionized in a solution of Pbl2? O O O adding a solution of NaNO3 adding a solution of Pb(NO3)2 addition of a catalyst adding a solution of KI changing the pH of the solutionarrow_forward
- The equilibrium expression for any weak acid can be written as HA (aq) + H20 (1) = A- (aq) + H;O+ (aq) 1. Write the K value expression based on the equation above (remember that pure liquids are not included in the K expression). This is given the special symbol Ka. 2. In this experiment, you will be using pH to find [H3O+]. The relationship is [H3O*] = 10-PH . For a pH of 7.4, find the [H3O+].arrow_forwardThe OH concentration in an aqueous solution at 25 °C is 7.9 x 10-3. What is [H]?arrow_forwardWhat is the concentration of ammonium at equilibrium if the initial concentration of ammonia, NH3, is 0.64 M and the K dissociation constant is 1.8 x 10-5 M? NH3(aq) + H₂O(1)→ NH+ OH-¹ 4(aq) (aq)arrow_forward
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