Question

Calculate the concentration of all species in a 0.235  molL−1 C6H5NH3Cl solution. (Kb for C6H5NH2Cl is 7.50×10−10)

Answer 1

C6H5NH3+ in aq.media disociate as shown below Concnetration of C6H5NH3+ = 0.235 M C6H5NH3+ is a weak acid So it ionize partially in aqueous solution to prodcue C6H5NH2 and H+ or H30+ The Ice table for weak acid is shown below. Where x is the amount of acid dissociated. <-------> + C6H5NH3+ (aq) 0.235 Initial Change Equilibrium C6H5NH2(aq) 0.00 + X (x) H+ (aq) 0.00 + X 0.235 -X = Weak B asedissociation constant (Kb) We know ka * Kb Then ka = 1.00E-14 7.50E-10 Kw Kw / Kb 1.00E-14 1.33E-05 = 7.50E-10 ka = M (C6H5NH2] [H+] Equilibrium expression (ka) = 5% approximation (C6H5NH3+] * 100 X* X Initial Conc 1.33E-05 0.235 -X 1.77E-03 - * 100 Since Ka is small, so x can be neglected in Denomirator 0.2350 X^2 = 0.753 % 1.33E-05 = 0.2350 0.753 < 5% X^2 1.333E-05 * 0.2350 3.133E-06 3.133E-06 11/2 1.770E-03 From ICE table X = [H+] or [H3O+] = 1.770E-03 The relation between pH and H3O+ is, pH = -log [H3O+] pH = -log [ 1.770E-03 ] = 2.752 so pH of weak acid = 2.75 From ICE table (C6H5NH3+] = = (0.235 - X) 0.23500 0.233 M - 1.770E-03 [C6H5NH2) = = [H3O+] or [H+] = X 1.77E-03 M = 1.00E-14 Tonic product of water Then [OH-] = = [H+] * [OH-] Kw / [H+] 1.00E-14 5.65E-12 M 1.77E-03 [CI-] = 0.235 M