Calculate the pH at 0mL, 5mL,...40mL for a 10.0mL aliquote of 0.100M Na3AsO4 (weak base) titrated with 0.100M HCl
pKa1 = 2.25
pKa2 = 6.77
pKa3 = 11.60
Thank you
First you should write equations to know where we are in the titration.
At the beginning, we have the hydrolysis of Na3AsO4.
AsO4^-3 + HOH ==> HAsO4^-2 + OH^-
k3 = [(HAsO4^-2)(OH^-)/(AsO4^-)]
Since (HAsO4^-2) = (OH^-) and you know (AsO4^-3), you can calculate (OH^-) and from that (H^+) and pH are available.
Next. Determine the points for the first, second and third equivalence points.
You have 0.010 L x 0.1 M AsO4^-3 or 0.001 mols. The first equivalence point then, titrating with 0.100 M HCl, will occur at 0.010 or 10 mL HCl. Therefore, at 5 mL we are half way to the first equivalence point. That means this reaction.
AsO4^-3 + H^+ ==> HAsO4^-2 + H2O
Use the Henderson-Hasslebalch equation.
pH = pk3 + log (base)/(acid)
AsO4^-3 is the base. HAsO4^-2 is the acid. Calculate the concentrations of each and solve for pH.
At the first equivalence point, you have H2AsO4^-2.
pH = sqrt(k2k3)
Continue in thi same fashion for each aliquot added. At the second equivalence point, pH = sqrt(k1k2)
At the third equivalence point, all of the AsO4^-3 is gone and replaced with H3AsO4. Don't forget to account for the dilution of the HCl as aliquots are added.
Post your work if you get stuck. I expect you can ignore the quadratic equations and make some approximations without much damage to the numbers.
Thank you DrBob222, but I'm really really slow. You wrote:
"k3 = [(HAsO4^-2)(OH^-)/(AsO4^-)]
Since (HAsO4^-2) = (OH^-) and you know (AsO4^-3), you can calculate (OH^-) and from that (H^+) and pH are available."
That means I do this, right?:
K3 = (x^2)/(0.1M of AsO4)
But how do I get K3 from pK3 if I don't know the concentrations of acid and OH^- yet? I see this formula in my textbook:
Ka = ( [H+][A-] / [HA] )
but I'm not sure what they mean by H+ A- and HA! I'm sorry.
The notation in your text and the notation I wrote (and you rewrote) is the same. The text means as follows:
(H^+) is hydrogen ion.
(A^-) is the anion. In your case this is the HAsO4^-2.
HA is the original acid. In this case it is HAsO4^-2. Where does pK3 come from? It is
pk3 = -log k3.
11.whatever = -log K.
Solve for K.
You should get about 2.51 x 10^-12
So this will end up as
2.51 x 10^-12 = (X^2)/0.1 and solve for X. You have let that be (OH^-). I would take the negative log of that as
pOH = - log(OH^-). Substitute the (OH^-) and obtain pOH. Then remember that
pH + pOH = pKw
pKw = 14. You will have pOH. Solve for pH. I hope this helps you get started. By the way, since pK3 is so small, the OH^- from H2O might have some effect but I have ignored that. Remember that H^+ = OH^- in water solution and that is 1 x 10^-7M so if your answer is near that number, then you may not be able to ignore the OH^- from water. That makes it a much more difficult part of the problem.
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