Question

1. Calculate the pH, pOH and percentage dissociation of a 0.20 M aqueous solution of the weak base, methylamine, CH,NH. The K. for CH,NH, is 3.6 x 10. (partial answer: pH= 11.93) 2. Calculate the pH of 0.75 M (CH3)2N. Ko for (CH3)2N is 7.4 x 10-(Answer pH=11.87) 3. Calculate the pH of 0.10 M NH.NO, K, for NH, is 5.6 x 10-20. (Answer pH=5.13)

Answer 1

The solution of the first part is given below

given [LU NH-020m 2 The reaction will be . (H₂NH₂ + H2O (H₂NH3 + + OH- atto 0.20 Chang X final 8.20-2 Now, Ko= [ (13 NH138] [or] [Cho Ng K6= 3.6*10-4 (ginen 3.6410-4 = [2] [2] 0. 20-2 Assuming 0.202 22= 0.72x 10-4 = 0.84410-2M x = 0.0084m sie [on] = 0.0084 Now Plon}= -log [ 0.0084) plov = 2.075 PH= 14-Plon) = 14.2.075 pH = 11.925 - 11.93 We know that kw=lka x kb. ka= Kw = 1x10m Kb 3.68104 Ka e 0.278 10-10 Now percent dissouation is given by. Amount dissociated x100 Initial anaud - [CH3 NI3] x 100 = 0.0684x100 = 4.2% [MN) 0.20

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