Question

Pyridine is a weak base with a Kb=1.7 x 10^-9. If 0.300 moles of pyridine is added to 1.00 L of water, what will be the equilibrium concentrations of the species below?

a. [Pyridine]

b.[Pyridine-H+] (the Pyridium ion)

c.[OH-]

2. what is the pH of the solution in problem 4 above?

Answer 1

Ko for Pyrimidine (Py) - 2.7450 Now 0.3 moles of l is added to tlt of water. [Py] = 0.3 (M) Keq [Py Ht] [ort] Now dissociation of ly is as follows, Py + H₂O & Py Ht + Ott- At aquidibrium [P, 7 [440] Keg. [Hz] Kb & [PyHt] [w] [PyJ Now let dissociation Constant a Now, let degree of dissociation of Py be $2. Hence at equilibrium, a Р. «Н,0 = Рунт. Он at equilibrium! 0:3 (1-2) 0.34 0.34. Kdo 0.36.x.083x 013 (1-2) rassuming ais Kao xe JK3 .. Every less J 0.342 03 7.538 LO [PyH 1 ) Hence at equilibrium, - 7.53 x 10X0:3 (M) Geote") - 2.259 * 205 (M) Ans: b 0.3 (1-7.53X105) (M) = 0.299971(M) Ans: a [Py] =

Now at Consider such low concentration, we should also dissorintion water too. H+ + OH- & Ht H₂o Kwa [H] [ou] 2014 Now song from Py contribution of [on ] > 2.250 x L (M) if disstei ionization of water produces x (M) on, *(x + @ 2.259 4108) to. >> me? + 2.289 XL6% x 10" z o » 2.259 x 69. Xotete comme venyny then, - O +4 2:289* 79110 (M) Hence overall [on ] at equilibrium 2 21283 xós + 4,43 x200 (M) . (If we consider disrorintion of water) If we don't consider, dissociation of water, then [ot], at equilibrium, & [PylHt Icg. 2.259 x 66% (M) => Ans e 2. Hence pH, 14 - POH .. + 4. 43420to) • 14 + log (2.259 20 79.357 Without considering dissociation of water. 14+ log (228 2.259 * 205 ) .9.35

So in this case calculation considering the dissociation of water was redundant,as for both cases pH values are almost very same.

But remember,for lower concentrations of weak acid or base,we should better check the dissociation of water too.