Question

Calculate the pH for each of the following solutions: a) 0.10 M NH₃ (K_b = 1.8 x10^-5) b) 0.050 M base with (pK_b = 6.13) (Can you please go step-by-step like Olsen Twins age 8 are trying to follow - thank you).

Answer 1

Given that concentration of NH3 = 0.1 M Kb of NH3 = 1.8 x 10-5 The reaction of NH3 in water: NH3 is a base, it reacts with water to form NH4t and OH-ions M NH3 NH3 + H20 + H2O — NH, NHA OH Initial 0.1 x Change + x Equillibrium 0.1 - x K, = X.X_=1.8x10-5 10.1-x x? +1.8x10-x-1.8x106 = 0 by solving the quadratic equation, we get, x = 0.001333 :: [OH-]=0.001333M POH = -log(0.001333) = 2.88 pH = 14 – 2.88 = 11.12

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Given that concentration of B = 0.050 M pKb = 6.13 Kb = 10-pKb = 10-6.13 = 7.4131 x 10-7 The reaction of B in water: B is a base, it reacts with water to form BH+ and OH-ions M B + H20 BH* Initial 0.050 Change -* Equillibrium 0.050 - X Ky = = 7.4131x10-7 x+ +(7.4131x10-?)x-(3.70655x10-4)= 0 by solving the quadratic equation, we get, x=0.0001922 :: [OH-]=0.0001922M pOH = -log(0.0001922) = 3.72 pH = 14 – 3.72 = 10.28