Question

Ammonia react with water as shown in the following equation: NH3(aq) + H2O(l)  NH4+(aq) + OH-(aq) Kb = 1.77 x 10-5 and Ka = 5.65 x 10-10 a) Calculate the pH of the 0.030 M NH3 solution (Use a labeled ICE table) b) Calculate the pH of the solution at the equivalent point (when equal amount of acid is added. Assume the volume doubles). (Use a labeled ICE table) c) Over which pH range (actual numbers) do you expect NH3(aq)/NH4+(aq) to behave as a buffer?  

Answer 1

NH+ (aq) + OM +0.0300 +0.0300 oH lagi OM +0.0302 to-0302 Cal Given Reaction - NH (aq) + Hoc ] - Initial 0.00M change -0.0304 0.30(1-0) Equilibrium Ka = Acid dissociation K = Bare dissociation - [NH+ oH 1 Kb = [ NH] 5.65810 10 1.77x10s Constant = Constant = (0.030212 0.030 (1-2) NH 0.03042 (1-0) l-an [oo weak Base] K = 0.0302² 91 Degree of dissociation x = kb 1-77x105 - 2.43X102 0.030 O-030 a 0.0343 TOH] = 0.030K = 0.030 x 0.0243 = 7.298104 poh = -log[or] = -log(7.29 X164) POH = 3.14 PH = 14-3-14 = 10.86 IPH = 10.86] Answer

(b) No 109) + HCl(49) NHA claqi Titial (0.030 Vlm-mole (0.oor ommole Change to.030 V/m-mole (60030V) + (0-osovim mole Equilibrium Ommole om.mole (+0.030vlom-mole Suppose 'V' is the volume of NH solution as well as HCl solution. Total volume of solution = av mL - No of m-moles of NH₃(aq) = 0.030MX V ml = 0.03 Vm.mole - No of m-mole of Hollag = 0.030 V m.moe. & Concentration of NH dlag) = - (0-030V)mmel 0.015M PH on Hydrolysis of salt of weak base and strong acid PH = √ { pkw -Pko - loge] pk = -log kt = -log (1.77x105) - pkb _ 4.75 I PH = √ {14 - 4.25 -logo-dism] py = 5.536) Ansion N Hadd(aq) + Holl) = NHylaq) + Hellas Til for a Buffer, PH range is from pkati to ka = 5-65x100 = Pka = -log(5-65x150 = 9.24 PH range = (9.24-1) to (9.24+1) PH range = 8.24 - 10.24