Question

Calculate the concentration of all species in a 0.225  mol/L C6H5NH3Cl solution. (Kb for C6H5NH2 is 7.50×10^-10)

Answer 1

The objective is to calculate the contuoti all species in 0.295 mol IL C& H5 NHEC Solution . - Given that kb for CCHE NH = 7.50 x 10 will a C CHg NH3 d = 0.225 molle o CG Hog Ntzel is an oionic compound. So, ut give 100% dissociation with Coty w Holante ion and Tona so, calz - [ Cg Hs Miltigt].m, 00925M ka x kb = kw = 1014 ka = 1014 kb = 0.133 x 10 initial 7.50 x180 Ka 1.33 x 0

Now , co Hg NH3 + H₂ot OG Hg No High + Ho 0.225 Initial -XM + M . change + 2m + 5M 일이 Equillirium (0.995 – )M aka : { CGH5 Ng ] [ ti3 ota [CHANHS) 1.93 x105 2. 9 . Ben (0.225-64] [6.225-23 +1 1.93 x 105 [0.295-x] - = 0.299 4167-f3310*cm at 1.93 10% Q - 0.299x.10+0. . a 1.6869 x 10 M son [H3O+ 3 = 4m 1.6865 x109 m [C6H5 NH3 ] = AM = 1,6865 x 163M [cits NH] = (0,225.-8) = [0.225 - 1.6865 x103] 0.2133 M :

[H+] = CH3 049 = 16865 x 18194 [H] [OH"] = 1014 COH-] = looked = 0.593 xlo M 1.6865x183 = 5.99 x 1012m [H] 5.95 * 100 ml Then Equilli bocum Cncentrations, (c] = 0.22M [ C6 Hg NH3 ] = 0.21M [CE H5 N Hz.] - 17 8163 M [H3O+] = [OH] = ax 10² M 5.93 x 162m