How to Calculate Kb and Find Its Value

[nautica]

I know this is simple but its been 2 years since gen chem and my brain is fried now. So, if someone could get me going in the right direction I would appreicate it.

Consider the reaction

(CH3CH2)2NH2+ ------- (CH3CH2)2NH + H+
Ka = 1.17x10^-11

Write reaction representing Kb and find the value of Kb

If I remeber right the negative log of the Ka will give me ph and I can subtract that from 14 and then reverse it to get the Kb. Is that right?

As far as the equation goes, I am not exactly sure what to do.

Thanks
Nautica

 

[Pyrrhus]

If i remember from my freshman chemistry Ka*Kb = Kw (Ionic Product of Water).

 

[movies]

The negative log of Ka will give you the pKa, which should be related to the pKb by this equation:

pKa + pKb = pKw = 14

which is derived from the equation Cyclovenom posted. Kw = 1 * 10^-14, by the way.

I think that the easiest way to solve it would be using the equation Cyclovenom posted though and just solve for Kb and plug in the numbers.

 

[ShawnD]

If I remeber right the negative log of the Ka will give me ph and I can subtract that from 14 and then reverse it to get the Kb.

Sounds about right.

pH + pOH = 14
ka * kb = $10^{-14}$

 

[nautica]

Cool.

So what do you think about the expresion?

 

[movies]

The reaction, you mean? The Kb equilibrium is the same as the Ka equilibrium, just in reverse.

 

[ShawnD]

Reversed, just like he said. I get a Kb value of 8.55x10^-4

 

[chem_tr]

I think the above discussion will work only if the medium is of an aqueous one. In acetic acid as the solvent, we cannot use 10^-14 of course. Please keep this in mind.

Also, please let me remind you that 1,17.10^-11 value represents the original, i.e., not protonated species, since -log value is about 11,5, this is basic. I think nautica should write the K_a value just below the unprotonated species, not the protonated one. In my opinion, the value that ShawnD found is the K_a of the protonated compound.

Regards, chem_tr

 

[movies]

Chem_tr is right that this only holds true in aqueous solution.

The Ka value corresponds to the entire equilibrium, not just one species in it. It is defined by whether or not the molecule is acting as an acid or a base, in this case the reactant side (left) has a molecule that is acting as an acid, hence you get the Ka when you figure the concentration of the product divided by the concentration of the reactants. The reverse reaction has a molecule acting as a base, so if you were to rewrite the equation flipped around from what is posted in the first message you would be dealing with a Kb equilibrium if you took products/reactants.

In summary, K applies to any equilibrium. Whether it is Ka or Kb depends on whether or not the molecule in question is acting as an acid or a base. So in this case, the Ka given is the Ka for the equilibrium drawn.

BTW, I get the same answer as ShawnD for the Kb.

 

[ShawnD]

In summary, K applies to any equilibrium. Whether it is Ka or Kb depends on whether or not the molecule in question is acting as an acid or a base. So in this case, the Ka given is the Ka for the equilibrium drawn.

I'll expand on that a little. Like you said, Ka or Kb means if it's acting like an acid or a base for the given equilibrium; it does not actually indicate if an equilibrium will be acidic or basic.
Example:
$H_3PO_4 \rightarrow H_2PO_4^- + H^+$

In the above example, going from H3PO4 to H2PO4 is the Ka value, going from H2PO4 to H3PO4 is the Kb value. Although H2PO4- is the conjugate base, it's still an acid.

 

[nautica]

So the Kb reaction would be?

(CH3CH2)2NH + H+ ------- (CH3CH2)2NH2+

 

[movies]

Yes, that's correct.

 

[chem_tr]

Oops! How did I make that mistake? Of course K_a points to a dissociation reaction of an acid. You are right, all of you.