Question

1. Calculate the pH at 25°C of a 0.757 M solution of a weak acid that has Ka = 9.28 x 10 2. Determine the Kb of a weak base if a 2.50 M solution of the base has a pH of 9.595 at 25°C. 3. The overall dissociation of malonic acid, H2C3H2O4, is represented below. The overall dissociation constant is also indicated. H2C3H204 = 2 H+ + C3H2042- K = 3.0 x 10-9 To a 0.025-molar solution of malonic acid, a strong acid is added until the pH is 1.5. Calculate the C3H2042-) in the resulting solution. (Assume the change in volume is negligible.) The value of the second dissociation constant, Kaz, is 2.0 x 10-6. 4. Determine the pH of a 2.25 M solution of pyridinium nitrate (CsH5NHNO3) at 25°C. [Pyridinium nitrate dissociates in water to give pyridinium ions (CsHsNH), the conjugate acid of pyridine (Kb = 1.73 x 10-9), and nitrate ions (NO3" ).)

Answer 1

1. Let the weak acid is HA. HA - 4t & Ä Initially the Gemeente concentrata is c= 0.757M. If degree of dissociation is a then the concentration of HA will be after dissociation - CHA] - (1-JC and concentration of it and l will be [H + ] = [A] = d.c. so [H ] = d.c. pH = -log[4+] = -logd - Loge. - - Log (ac) ka= [HT] [A] CHA] - 9.28 X70-6 (2C). (d) x2 (1-0).c (1 x) 9.287106 1.22 10 0:757 > 2 + (1.22*10-5) - (1.20185) = 0

so d = - (1.224 10 5) + 1100) + J (1.22*10 J (1.22 4 10 5)2 + (401. 22x105) &- can not be negative. d = -1.22 x 105) + J (1-22 x 10 52 + (48 1.224105) 3.48 810 3. pH = -log(&c) -- 19 (3.48 % 16 % x 0.757) = 2:58 8. 2.58. The pu of the solution will be