Finding Ka from pH Titration Graph


by BayernBlues
Tags: graph, titration
BayernBlues
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#1
Feb16-08, 12:16 AM
P: 66
1. The problem statement, all variables and given/known data

Need a little help on a lab. I have all the data for a titration of a weak acid with NaOH which includes the pH of the acid for however many mL of NaOH added. I've plotted a pH vs V NaOH added graph. From that, I plotted a ∆pH/∆V vs V NaOH added graph. From this, I found my equivalence point but the lab is asking me to find Ka for the unknown acid at 0%, 20%, 60% etc titration points where 100% is the equivalence point. From this, I have to identify the acid (it's either acetic acid, monochloroacetic acid, dichloro acetic acid etc.

I've found my equivalence point BUT I don't know how to find the Ka for the unknown acid from this equivalence point.

2. Relevant equations

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

3. The attempt at a solution

[NaOH] is given as 0.099 M (NaOH)
I'm guessing [HA] = CBVB/VA
Also I'm guessing that [H+] = 10^(-pH) where pH is the ORIGINAL pH of the mixture?
I'm unsure on what [OH-] would be.
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physixguru
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#2
Feb16-08, 12:56 AM
P: 339



Titrations are often recorded on titration curves, whose compositions are generally identical: the independent variable is the volume of the titrant, while the dependent variable is the pH of the solution (which changes depending on the composition of the two solutions). The equivalence point is a significant point on the graph (the point at which all of the starting solution, usually an acid, has been neutralized by the titrant, usually a base). It can be calculated precisely by finding the second derivative of the titration curve and computing the points of inflection (where the graph changes concavity); however, in most cases, simple visual inspection of the curve will suffice (in the curve given to the right, both equivalence points are visible, after roughly 15 and 30 mL of NaOH solution has been titrated into the oxalic acid solution.) To calculate the pKa values, one must find the volume at the half-equivalence point, that is where half the amount of titrant has been added to form the next compound (here, sodium hydrogen oxalate, then disodium oxalate). Halfway between each equivalence point, at 7.5 mL and 22.5 mL, the pH observed was about 1.5 and 4, giving the pKa values.

In monoprotic acids, the point halfway between the beginning of the curve (before any titrant has been added) and the equivalence point is significant: at that point, the concentrations of the two species (the acid and conjugate base) are equal. Therefore, the Henderson-Hasselbalch equation can be solved in this manner:



Therefore, one can easily find the acid dissociation constant of the monoprotic acid by finding the pH of the point halfway between the beginning of the curve and the equivalence point, and solving the simplified equation. In the case of the sample curve, the Ka would be approximately 1.7810-5 from visual inspection (the actual Ka2 is 1.710-5)

For polyprotic acids, calculating the acid dissociation constants is only marginally more difficult: the first acid dissociation constant can be calculated the same way as it would be calculated in a monoprotic acid. The second acid dissociation constant, however, is the point halfway between the first equivalence point and the second equivalence point (and so on for acids that release more than two protons, such as phosphoric acid).
BayernBlues
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#3
Feb16-08, 12:16 PM
P: 66
yep, my first graph looks like that and I have the equivalence point from the derivative of that graph. I'm supposed to use a different mothod to find Ka though using this equation:

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

I'm unsure though how to find what [H+] and [A-] would be at the equivalence point. I can find [HA] from the volume of base added though.

chemisttree
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#4
Feb18-08, 02:33 PM
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Finding Ka from pH Titration Graph


From your equation (Ka=[H+][A-]/[HA]) you can look at the graph before you add any titrant. The pH of the solution before you titrate will give you the concentration of [H+] (and the concentration of [A-]). That concentration ( [H+] ) arises by deprotonation of some of the [HA].

Soooo, if you know that the concentration of [HA] at equilibrium is actually [HA] - [H+], the expression for Ka becomes, Ka = [H+]^2/([HA]-[H+]). Substitute the value of total [HA] that you obtain by completely titrating the acid and the pH you measured before you began to obtain Ka.


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