102. Calculate the percentage of pyridine (C5H5N) that forms pyri- dinium ion, C-H5NH+, in a 0.10-M aqueous solution of pyri- dine (K, 1.7 x 10-⁹). =

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.)

Potassium hydroxide (KOH) is a strong base. Dissociation of potassium hydroxide follows the reaction equation given. KOH(s) --H20--> K+(aq) OH-(aq)   a) Write down the reaction equation for a neutralization reaction where potassium hydroxide (KOH) reacts with sulphuric acid (H2SO4) producing potassium sulphate (K2SO4) and water (H2O). Balance if needed! b) Calculate pH for the solution where 0,2 mol of KOH is dissolved into water so that the total volume of the solution is 0,5 L.

8. Calculate the equilibrium concentration of each aqueous species (H2A, HA– A2–, H3O+, and OH–) in the solution. Assume a temperature of 25°C. H2A: pKa1 = 2.05 and pKa2 = 6.22a. 0.10 M H2A b. 0.10 M Na2A