Question

2. Cara tepat 02 mocasion with oven to k 170x2109 2. Calculate the pH of a 0.22 M solution of NH F given the K, is 1.79 X 105

Answer 1

Given, 0.22M solution of NH₃F Kb = 1.79x10-5 Reactions envolved: NH4F - NHYT & F NHL + H₂O & NH₃ + H₂O + Initially: 0.22M OM OM (Dissociation reaction) (Ionization reaction) 0.22 M OM OM ta ta At Equilibrium: (0.22-2)M RM RM Now, for a conjugate acid-base pair, we have Ka.Kb = Kw = 1x10-14 ka = Kw = 1X10-14 = 0.558 x 10 9 = 5.6810-10 Kb 1.79x10-5 Also, ka = [NH3] [H3O+] = [ NAut] (22) (2) (0.22-x)

Now, lets check if we can ignore the factor 'a' in (0.22-27). So, il 400 times ka< Duitial concentration then we can drop a im (0.22-3). So, 400 x (5-6 X 10 -0) <0.22 clearly, it is indeed yes. i.e. even if the dissociation constant is multiplied by 400, then it is much lesser than the initial couscentration. which means in (a) (1), the denominator is very large. 0.22-2 So, 0.22 >>> :.0.22-2 a 0.22 Now, ka = x2 = 5.6x10-10 0.22 = 55.6* 0:22 x 10-10 - 1•109 X10-S = [NM3] = [H30 +7 :. PH- Log. [+] = -log(1.109x10-5) = - [log, 1.109 t tos, 10-s -- [lofio 109 - 5 lxo'] -- [0.0449 - 5 - 4.9550 i ph of the solution - 4.9550 = 4.96
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