How do weak bases have Kb?

Question

If you think about NH3 in an acidic solution, it is accepting H+ in the water to form NH4 and release OH- (which is why in the addition of NH3 the solution becomes slightly more basic). I don't really understand how it has a "dissociation constant" (dissociation=separation) if it is actually accepting H+. Is the dissociation referring to the H+ that dissociates from water upon the addition of a base? Also, what is the unit of this dissociation constant/what is its relation to moles? For example, if I have 0.05 mol HCl, a strong acid that dissociates completely, I wouldn't add 0.05 mol NH3 to counter it because NH3 doesn't accept H+ at the same rate HCl releases H+ correct? Which is also why the equivalence point of this reaction is 5.13 instead of 7, or neutral. So upon the addition of equal volumes+concentrations of NH3 and HCl, there is an excess of H+ still in the solution. Then how does the mole relate to the pH scale? Since it is logarithmic, 10^-1 is pH 1. So what is this? Is is 10^-1 particles, atoms, or moles of H? 

I would really, really appreciate it if someone could help me out (even if you can only answer parts of the question). I'm having a lot of trouble understanding this, and I'm not a huge fan of plugging into equations so I want to be able to picture what's going on. Thanks!

Nolan · Accepted Answer

The "dissociation" constant for all bases is, essentially, the dissociation of water. Everything to do with acids and bases is pretty much defined by how acids are measured.

All equilibrium constants (Kc values) are unitless (in general chemistry). The explanation that my textbook gives (Burdge, Chemistry 2nd) is that each concentration is divided by a reference concentration (or a partial pressure, for gasses), which cancels all of the units without changing the values. My professor explained to us that equilibrium expressions in gen chem are the somewhat simplified results of a larger series of quantitative analysis that is used in analytical chem.

The last part of the first paragraph is a question that buffer solutions deals with, Like you said, HCl dissociates completely. The chloric ions will be spectators, and can therefore be ignored. Adding protons (H+) to a weak acid or base solution will stress the solution's equilibrium and cause it to shift in the direction that minimizes that stress. All of the added protons will be used up (as long as there are enough A- ions to soak them up), and the equilibrium will be reestablished at it's new point.

The mole is used to determine the concentration of stuff in solution (through the Molarity equation), so it's related to pH through it's use in determining concentration. I think that you're asking about this because titration problems tend to use moles in the first part of solutions, though. You can do that because the small amounts of acid (or base) that you're adding to the titration shouldn't change the volume, and therefore shouldn't change the Molarity.

pH itself is defined as the negative log of the hydronium (or proton) concentration, though. It's simply a way to squeeze a HUGE range of numbers (from 0.1 through 0.00000000000001) into a more manageable range from 1 through 14 (note though that the pH and pOH scale is actually infinite [acetic acid can have concentrations above 1.0 M and below 1.0x10^-14 M), it's just unusual to see any values outside of the 1 through 14 range. so, ultimately, pH tells you the Molarity of the acid.

skipper · Answer

You would add 0.05 mole of NH3 to neutralize 0.05 mole of HCl. You are correct in stating that the pH would be less than 7.0, This is because-

NH4^+1 + H2O <--> NH3 + H3O^+1

The conjugate acid (NH4+) of the base (NH3) reacts with the base (H2O) to form the acid (H3O+).

Kb = [NH3][H3O+] / [NH4+]

If you mixed a solution so that it contained 0.05 mole each of NH3 and HCl, some of the NH4^+1 that is formed would react with the water to yield H3O^+1. The value of Kb for NH3 is 1.78x10^-5 you

Kb = 1.78x10^-5 = [H3O+]^2 / (0.05 - [H3O+])

[H3O+] = 9.43x10^-4

pH = 3.03

Weak acid:

HA + H2O <--> H3O^+1 + A^-1

Ka = [H3O+][A-] / [HA]

Weak base:

A^-1 is the conjugate base of weak acid HA

A^-1 + H2O <--> HA + OH^-1

Kb = [HA][OH-]/[[A-]

Ka x Kb = [H3O+][A-]/[HA] x [HA][OH-]/[A-]

cancel like terms to get-

[H3O+][OH-]

Ka x Kb = [H30+][OH] = Kw

2 H2O <--> H3O^+1 + OH^-1

B + H2O <--> BH^+1 + OH^-1

Kb = [BH+][OH-]/[B]

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How do weak bases have Kb?

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