Question

Quest. 4 (15 pts). Ammonia react with water as shown in the following equation: NH3(aq) + H2O(l) NH4+(aq) + OH(aq) Kb = 1.77 x 10-4 and Ka = 5.65 x 10-10 a) Calculate the pH of the 0.030 M NH solution (Use a labeled ICE table) b) Calculate the pH of the solution at the equivalent point (when equal amount on 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

dissociates Let, Nt3 dissociates Opto équilim is xml L (4) (6) NH3 is a weak acid, which in water as: NH + H2O Nty + + oH T: 003 - -a - E. (0.03-2) – x. x .. Equilibrium constant K6= Inty deg [04 Jeg [Nitleg 178X20-S3 The second ->) 177x10-5 - the vahe of kь и .// هم So, dissociation of NH3 will bevery small. 0.03-x 003 - 1.77810 - 5 x 0103=n 54 51 3/X10-7=21 . : S. ) X10-7.)” * = 7.3 x 10-4 ms Colleg= 7.3 x 104 polt=-Log Lottideg = -log ( 3x10-4) (? in 10 33.94 .: PH+ Polt=14 & PH=14- polt = 14- 3.14 = 10.86 .: PH of 0.030M NH3 solutionis 1086 (6) At equivalence point all the NH3 converted to whet .: Volume in double. So at eanivalence point In Hut) = 4x [Na] ayx 0.00 m -0.015OM

(6) Nyt will be hydrolysed. N. Hut in a conjugate seid of weak bare in & Ho - Nyt NH3 + Hoot I: 0.0750 - ty 7 ci -Y E: 10.0lsory) - – 9 up to reanin Exet, change of Intut by A7 Equilibriuen constant Ka [Niegt the deg na [ Nikt er u very small. گل 000,ی - =) 565 x 1810 - YXY Toolsity) 5) 5.65 x2000 5°45X/0° 2 y: 5:65x 70*70X 010150 . .- (5.65x10 - 10x 0.0150) nooisy 7- 2.9x106 test) = 2.9x186 en PH = log hot an log (2.9x106) - .. PH at equivalence point acill be 5.54 070 V 10 5.54

(c) & suffer shower buffering copper ty when pt sarge iu pka-) to gkat now, pka of Nity --Lota ng (565x600) - 9.25 NH3 (as) /NHut laq acill be have as baffer when ptrange in 9.25-1 to 925+1 ; l. 8.25 to 10:25