Question

Activity #2 Calculate the pH of the following basic solutions. 1. Calculate the pH and pOH of a 0.200 M Ca(OH)2 2. Calculate the equilibrium concentrations for [OH-], [HB+), and [B], the pH and pOH for a 0.200 M pyridine solution. (CsH5N) Kb = 1.4 X 10-9 3. Calculate the equilibrium concentrations for [OH-], [HB], and [B], the pH and pOH for a 0.200 M methylamine. (CH3NH2) Kb = 4.4 X 10-4 4. Calculate the equilibrium concentrations for [OH-], [HB+], and [B], the pH and pOH for a 0.200 M ammonia solution. (NH3Kb = 1.8 X 10-5 Use your calculation results to complete the following table POH pH Kb Base Ca(OH)2 CSHSN CH3NH2 NH3 What is relationship between the Kb and the pOH and pH of the solution? Rank the following bases in terms of their strength from weak to stronger.

Answer 1

Ca (OH)2 Ca2+ + 2 OH- Imol La COH) gives a moles of OH 10.200m Ca (Oly gives 0-200x2=0.400m otions [OH-] = 0.400m pOH = -log [OH-] = -log (0.400) = 0.398 pH + pot = 14 + pH = 14- pot = 14-0.298 pH = 13.60

pot pH 1 Base Kb CaCOH)₂ 10.398 1.48109 CHN 14.78 9.22 CH₂ NlHz 2.03 11.97 4.48104 NH3 2.72 [ 11.28 11.28 1.8x105 Ky and pott are inversely proportional to each other kb and pH are directly proportional to each other Ranking : CS HSN <NH3 < CH₃NH₂ < Ca (OH),

3 pOH = 2 pkb - 1/₂ log C = 2.027758664 pH = 11.9722 [OH-] = [HB+] = 102.027768664 = 0.0093808m [B] = 0.200 -0.0093508 = 0.190619 m 4 pH = 1/₂ (- log (1.8x105)) } Log (0.200) = 2.722 pH = 11.28 [OH-] = [H8+] = 102.722 = 0.0018 967 m [B] = 0.200 -0.0018967 = 0.1981m.

2) - - CANH + G H₂ N + H₂O B BH + for a weak base, pot = 1 pkb - I loge pkb log kb = -log (1.4x109 = 8.653872 c = {B} = ( CJ HsN] = 0.200 m pH = (8.853872) - Log (0.200) = 4.7764 pH = 14 - pH = 9. 2236 COH-] = 104. TT64 = 1.67340105 m (pH = -log (OH)) [H8+] = [OH-] = 1.673 x 10 m [B] = c-x = 0.200 (1.673x105) = 0.199983 m