physical chemistry

[Lisa_g]

The weak base hydrazine. jas the chemical formula NH2 NH2

(pKb1 = 5.77; pKb2 = 15.05)

Write a balanced equations showing the first and second base dissociations, relating to Kb1 and Kb2 in water.

All ive been given is the above details and am confused on how to tackle this.
Any answer would be appreciated
Regards, Lisa

[Hix3r]

The pKb is a number which is -lg(Kb) of the solution. It tells you the concentration of the product or reagent when exactly half of them dissociated.

NH2-NH2 + H2O <--> NH3+-NH2 + OH-

NH3+-NH2 + H2O <--> NH3+-NH3+ + OH-

Anyway when [NH2-NH2]e = [NH3+-NH2]e (so they equilibrium concentrations are exactly the same), the first pKb is in action.

And when [NH3+-NH2]e = [NH3+-NH3+]e then the second pKb comes into play. Sorry to write 15 here I was not paying attention. I was wrong.

pKb is easily derived from Kb. pKb = -lg(Kb).
It is much used in the Henderson-Hasselbach equation, which is the main thing to calculate exact pH of weak acids, bases.

pH = pKb + lg([conc. of deprotonated molecules]/[conc. of protonated molecules])

So if you want to calculate Kb:
- take the pKb
- put ten to the powers of negative pKb
- so if pKb=2, Kb=10^-2

[Lisa_g]

your very clever mate, thanks alot but one thing thats still puzzling me is how do you get pH 15.05, i thought pH only went upto 14. thanks again

[Hix3r]

That is true. Truly sorry for the mixup. It is not pH, its pKb. It is only the negative logarithm of the Kb value. And the Kb (equilibrium constant or whatever) for the second -NH2 is about approximately 8.4*10^-16. This very little value tells us that the dissociation of the second NH2 is very little.





So these pictures tells us that K is (multiplied concentration of products on the power of how many the reaction produces) / (multiplied concentration of reagents on the power of how many the reaction requires)

Now think if there very little of the product (very small number) and lots of the reagents (very big number). Then if you divide a very small number with a very big number, you get an even smaller number possibly very very little. That is why this Kb is so small.

There is a trick example. What is the pH of a 10^-8 M HCl solution?
okay its a strong acid so you try to calculate pH: pH = -lg[H+]. HCl dissociates completely, therefore [H+]=10^-8 mol/l now is you take the logarithm and take the negative you get pH=8!!! Now wait a minute an acid is basic in a solution????

No it is not. Water dissociates too! But it is almost non-existent. In water there are always 10^-7 mol/l H+ and 10^-7 mol/l OH-. Because of this the pH will be 7. But if you add 10^-8 mol/l H+ to it, then it will be 1.1*10^-7 mol/l H+, and if you calculate the pH: pH= -lg[H+] you get 6.96. Which is the correct one. So at cases where the dissociation of the molecule is very small you have to calculate with the dissociation of the water too.

I told you that the pKb says what will be the pH when there are exactly the same amount of dissociated and non-dissociated molecules. It is true if the number of dissociated molecules would be larger than the water dissociation. So the pH is below 14, that can be calculated simply:
pKb = -lg(Kb) and pH = -lg[H+] so [H+] is exactly Kb at that particulary point. and it is way smaller than 10^-7. If we add them together:

pOH = -lg[Kb + OH-] = -lg[0.00000010000000084] wow
and then pH = 14 - pOH

and if you take the logarithm and multiply with -1 you get that the pH at the point where [NH3+-NH3+] = [NH3+-NH2] is exactly 7.0000000036480736326654061165255. So it's a little over seven. This is because only few molecules actually dissociate.
Therefore pKb is only telling us the pH at that point where the dissociation of water does not interfere, so its much bigger than that.
If you calculate pH from the first pKb then you will notice that the pH there can be 8.49 or so. (Depends on concentration. I used 1M solutions.) So be careful with this! I messed up too.

To make it clear in your head. There is pKa and pKb. And pKa+pKb=14.
pKa tells you the pH of the solution when exactly half of the molecules dissociated and it is more than the water's dissociation. pKb tells you the pOH of a solution when the above mentioned conditions apply. You can easily calculate pKa therefore the pH at that state of the solution from pKb from the above mentioned equation. Don't confuse them like I did.

[Lisa_g]

you are so sweet and clever, wish i was clever like you.

how do i determine the values of the dissociation constants for both reactions kb1 and kb2, with the concentration of the base being at 0.2M and subsequently the pOH for both dissociations reactions in water

thanks for all your help

[Lisa_g]

since my last post ive been trying to work out the pH for kb1
what i got is

10-pkb1 = kb1
10 – 5.77 = kb1 = 1.70 x10 -6

1.70 x10 -6 = [ OH-]2
0.02

OH- = √ 1.70 x10 -6 X 0.02
= 1.84 x10 -4

pOH = - log [OH-]
= - log (1.84 x10 -4)
= 3.74

pH = 14 – pOH
= 14 – 3.74
pH = 10.26

dont know if thats right but can someone correct me if i've gone wrong any where

and i got 5.63 for my 2nd pH, is it suppose to be below or above 7

[Hix3r]

That seems allright.

Now for thew second I got 7.02 here take a look:

10^-15.05 = 8.91*10^-16

[OH-] = √(8.91*10^-16 * 0.02) (the NH3+-NH2 conc. is 0.02 at this point, because the second group will only start to protonate when the first one finished)

[OH-] = 4.22*10^-9

here we add the water's part (this is important here, water makes 100x more OH- here so it determines the pH):

[OH-] = 4.22*10^-9 + 10^-7 = 1.0422*10^-7

pOH= -lg(1.0422*10^-7) = 6.98

pH = 14 - pOH = 7.02

Now if you think about it. It can't go below seven because that would mean that the molecule actually increases the [H+] concentration in the solution, but it is a base right? But the Kb tells us that only small part of the molecules will protonate, so basically it is hard to get those groups to accept protons. Therefore we must increase the proton concentration to make them take up protons. And if [H+] concentration goes up, pH goes down. Actually the pH goes down but should not go below 7.

[Lisa_g]

yeh your right, i forgot to put the minus at the beginning, silly me

i've done everything now, thanks for all your help