[pdwnz]

Equal moles of ammonium chloride solution and ammonia solution are mixed.

Upon inspection of both reactions we see that for the hydrolysis of ammonium chloride that hydronium ion and ammonia are products and for the hydrolysis of ammonia in the second solution that ammonium ions and hydroxide ions are produced.

obviously being that the pH of the resulting buffer is slightly alkaline, it must be that a small excess of hydroxide ion is present...how is this possible given that the hydronium ion produced via the hydrolysis of the first ammonium chloride solution would mop up all of the hydroxide ions produced in the second reaction...according to the stoichiometry all species are 1 to 1...?

confused please explain. Thanks.

[Arkcon]

Draw these reactions for us in chemical notation.  You are over simplifying the dissociation reactions  -- particularly for ammonium chloride.

[pdwnz]

NH₄⁺  +  H₂O     H₃O⁺  +  NH₃

Ka = 5.8x10-10

NH₃  +  H₂O     NH₄⁺  +  OH-

Kb = 1.74×10-5

Kb 》Ka  is this why buffer is slightly alkaline?  If so, some  further elaboration with respect to the math, and further simple rationale, that explains the role of subsequent dissociations of the products with ever diminishing K values in the determination of the pH of 9.24 would be appreciated.

[pdwnz]

Another source of confusion comes when I use the Ka and Kb equilibrium constant values to calculate  the  relative concentrations of hydronium to hydroxide ion.

Sure more hydroxide is present as the Kb value implies the equilibrium lies further right for the hydrolysis of ammonia relative to the position of equilibrium for ammoniumm, but the derived OH- calculation for the first dissociation gives a pH much higher than 9.24.

Even taking into consideration subsequent shifts in the equilibrium position, I find hard to logic out the pH value for this buffer. 

[Babcock_Hall]

Have you tried to perform a calculation based on the information that you have?

[QuantumSufi]

Since the Ka for ammonium ion is very small, and also since there is much more ammonia present in the solution implying that the first equilibrium lies mostly to the left i.e. ammonium ions are present undissociated. And ammonia in the solution is contributed almost entirely from the ammonia solution. Using the equation
                                            Ka =   ([NH₃][H⁺])/[NH₄⁺]

for the first equilibrium, setting [NH₃] = [NH₄⁺] and using the definition of pH, you can convince yourself that the pH turns out to be 9.25.
You may want to look up the following page:
http://chemwiki.ucdavis.edu/Physical_Chemistry/Equilibria/Acid-Base_Equilibria/7._Buffer_Solutions

[pdwnz]

Okay thanks for the  help.

I think I am on track with this now mostly, still a little tentative with the exact derivation of pH of 9.24 and here is why.

As we know the pka = pH when [conj base] = [acid]

In a problem where the starting concentration of the acid is given, use of ka allows the calculation of the conjugate base and the same can be said when calculating the concentration of hydroxide using kb, and this is where my understanding begins to wane.

E.g. suppose the starting concentration of NH₃ = 0.5M 
kb = 1.74 × 10^-5 gives [OH-] = 2.95 × 10^-3M which equates to a pH of 11.44

Obviously this is too high and competing equilibria serves to lower the [OH-] to a level where pH = pka but some math and rationale to elaborate by someone in the know would help.

[Borek]

Best approach is to solve whole thing - start with equations describing all equilibria present in the solution (actually there are only two of them that matter), add mass/charge balances, then solve the system of the equations.

Not much different from what is done here:

http://www.chembuddy.com/?left=pH-calculation&right=pH-salt-solution

Just Ka (dissociation of the HCl, not of the NH₄⁺!) is so high it can be assumed to be infinite which means we can ignore it. That leaves us with two equilibria - water autodissociation and ammonia reaction with water.

Don't think in terms of ammonia Ka and Kb - whole system is described by either Ka/Kw or Kb/Kw, using Ka/Kb/Kw system just adds an equation that is not independent (ie it doesn't give us any additional information, as it can be expressed by other two). After all, Ka*Kb=Kw, so knowing any two we can calculate the third.

[Arkcon]

What you're doing is creating a buffer -- mixing a base and its acid salt.  That is defined as pH=pKa-log₁₀ (conc. base)/(conc. acid).  Now, since you're not making a practical buffer, but are instead interested in the relationship between Ka and Kb, maybe your text book (or more likely a more advanced textbook) can explain how the "workhorse" buffer calculation is derived from the other formulas you're trying to work with.