Hallo, I need help with a hard chem problem
A 30.0 ml sample of a 0.12M aqueous solution of hydrazine, N2H4, is titrated with 0.10M HCl. Given that the Kb of hydrazine is 3.0e-6, determine the pH...
(a) before the titration begins
(b) at the halfway point
(c) at the equivalence point
(d) at a point somewhere between the halfway point and the equivalence point
(e) at a point 30-50ml past the equivalence point
So far, I've solved a, b, and e, but i'm stuck mostly on c and d Q_Q a b and e don't require solving for the concentrations so they were easier
The henderson-hasselbalch equation isn't helping all that much right now since this doesn't give me a clear conjugate base for the hcl, so I was wondering if there was maybe something I could do there
I've been staring at this and trying to solve it for a while but I'm so bad at this halp
3 Comments • Newest first
[quote=ehnogi]Sure why not. I have time, and I can't sleep.
[/quote]
omg i love you so much right now
Suddenly I question how I aced chem..
Sure why not. I have time, and I can't sleep.
[b]Part A[/b] ---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
To find the pH before the titration begins, you need to find the relationship between the Kb and ( [OH-] [A-] ) / [HA] , where [OH-] = concentration of hydroxide ions , [A-] is the concentration of conjugate acid, and [HA] is the remaining hydrazine.
To do this, you need an ICE table:
N2H4 (aq) + H2O (l) --><-- N2H5 (aq) + OH (aq)
Initial 0.12 M ------ 0 0
Change -x ------ +x +x
Equilibrium 0.12 - x ------ x x
Observe that the [HA] is remaining hydrazine, N2H5 is your conjugate acid and OH- is just your OH.
Plug into this equation :
Kb = ( [H+] [A-] ) / [HA]
And you get the equation 3.0 x 10^-6 = (x^2)/(0.12 - x). Solve for x;
x = 6.0 x 10^-4 ,
where x represents your [OH-] concentration.
To find the pH of a solution, you can find the pOH of the solution and subtract it from 14.
To find the pOH of the solution, you find the -log[OH-].
-log[OH-] = 3.2
14 - 3.2 = 10.8.
pH = 10.8
[b]Part B[/b]---------------------------------------------------------------------------------------------------------------------------------------------------------------------------------
So you really only have to remember the "half-way point" as the "buffer zone". [b]In the buffer zone, pH = pKa. [/b]
This is derived from the Henderson Hasselbach equation pH = pKa + log([A-]/[HA]); at the buffer zone, the ratio of base to acid is almost 1 to 1.
Since you already know your Kb, we can find the Ka.
We know that Kw = 1.0 x 10^-14, and we also know that Kw = Ka * Kb.
Using this relation, we can isolate Ka by plugging in for Kb and Kw like so :
1.0 x 10^-14 = Ka * 3.0 x 10^-6
(1.0 x 10^-14)/(3.0 x 10^-6) = Ka
Ka = 3.0 x 10^-9
We can now find the pKa by taking the -log(Ka), which would be 8.52. It makes sense that the pH is still basic because we didn't reach the equivalence point, but it's less basic than before because we are titrating with an acid.
[b]Part C[/b] -----------------------------------------------------------------------------------------------------------------------------------------------------------------
The equivalence point describes that the moles of solution you started with equals to the moles of titrant added. With a strong acid and strong base, the pH is 7; with a strong acid and weak base, the equivalence will be acidic.
We need to find the molar concentration of H+ in our equivalence point (because there won't be any OH- left to be significant enough to change the pH); in order to do that, we need to first convert everything into moles.
Moles of hydrazine: 30 mL x 0.12 M hydrazine = 0.0036 moles hydrazine
Remember that moles hydrazine = moles HCl added.
Therefore there were 0.0036 HCl moles added.
Using M = n/V, we find that
0.10 M HCl = 0.0036 mol HCl / V
V = 0.036 L , or 36 mL.
We know that the volume of HCl added was 36 mL. This makes the entire solution at the equivalence point 60 mL. You completely disregard the moles of OH- that was initially in the solution because it is literally obliterated by the moles of H+ (0.0036 H+ - 0.000018 moles OH- is still around 0.0036 moles of H+), so just find the molarity of the N2H5+ in the total solution.
The following in moles:
H+ (from HCl) N2H4 N2H5+
Initial | 0.0036 | 0.0036 | 0
Change | -x | -x | +x
Final | 0 | 0 | +0.0036
^ We know that concentration of N2H4 and HCl are the same, and we know almost all of the N2H4 is being shifted over to make N2H5.
So we know that N2H5+ is the only thing affecting the pH now.
We can find the concentration of this :
Total volume = 36 mL + 30 mL = 66 mL.
[N2H5] = 0.0036 / 0.066
[N2H5] = .055
Now we need to determine the concentration of H+ through the ICE table again.
Remember that Ka = 3.0 x 10^-9, and Ka = [H+][A-]/[HA]
Therefore:
3.0 x 10^-9 = x^2/0.55 - x
x = 4.1 x 10^-5 = [H+]
pH = -log[H+]
pH = 4.4
Yup. This is why I hate chemistry. Coming up with part D...
[b]Part D[/b]-----------------------------------------------------------------
You don't use Henderssen Hasselbach for HCl because HCl dissociates completely. You use it for N2H5.
[b]Part E[/b] ----------------------------------------------------------------
You calculate the excess H+ concentration and use that to find the pH