Titration curves lab?

Question

okay Im not quite sure how to get the Dissociation constant. I have the rest of the values though.

I have 7.00 ml for the first derivative, 6.93ml for the 2nd. It then asks for the average which is 6.97ml

And then it asks for the molarity which is .100M We titrated a unknown acid with .100 NaOH

I have pOH = pKb + log [conjugate acid]/[base]

Im not sure which volume I would use to get the pOH because for some reason the lab instructor told this other guy to devide his average vol by 2 and use that volume to get the pH at that volume. And then plug that into the equation.

Im also not sure which values to use at conjugate acid and base

So can anyone help?

I believe the moles is .1000 Molarity of NaOH x .00697L average liters of NaOH = .000697 moles

okay I think it wants the acid in moles not NaOH, thats .1000 x .010 = .001 moles of acid

Answer 1

At the equivalence point the curve changes its curvature. Thus instead of trying to find the middle of the curve in order to determine the endpoint you can find it by doing the 1st derivative (the endpoint will be a peak) or the 2nd (at the endpoint it will be 0))
Since you are using experimental data, when you are calculating the derivatives you will have an error depending on the amount of base added at each step (the bigger the volumes, the bigger the error; remember that detivatives require by definition extremely small changes in order to be accurate). I guess your instructor told you to take the average of the two in order to minimize the error.

The equation is actually pH=pKa+log[conj.base]/[acid]. At half equlivalence point [conj.base]/[acid]=1 thus pH=pKa+log1=pKa.

If your equivalence point is at 6.97 ml then the half equivalence point is at 3.48-3.49 ml. Go to your pH vs V base plot, and find the pH value for that volume. For that value pH=pKa and thus Ka=10^-pKa

Answer 2

in view that its sturdy acid+sturdy base the product is H2O and salt, you be attentive to its going to be a one million:one million reaction, and you have the concentrations. Ba(OH)(aq) + HCl(aq) --> H2O(l) + BaCl(aq) *continually contain the equation, (aq) potential aqueous, fantastically lots something in answer the 1st ingredient is sturdy, the final ingredient could be pH=7. upload 3mL each and every ingredient not 5, the equivalence ingredient could be reached at 12mL HCl. one million/one million.25=0.8 and 0.8*15mL=12mL, so 0ml, 3ml,6ml,9ml, and 12ml. sorry beforehand for the loopy thank you to place in writing it, extra straightforward on paper for sure 2d ingredient could bypass; one million.5*10^(-3)mol OH - 3.seventy 5*10^(-4)mol H+ = one million.a hundred twenty five*10^(3-) OH /*makes slightly salt and water pOH= -log(one million.a hundred twenty five*10^(3-) OH / .018L = one million.2 pH = 14 - one million.2 = 12.8 comparable steps for the subsequent factors; ingredient 3 6mL HCl pOH=one million.447 pH=12.5 ingredient 4 9mL HCl pOH=one million.8 pH=12.2 ingredient 5 12mL HCl pH=pOH=7 * you may to function a ingredient like 11mL or graph it with .01mL increments, the reason at the back of that's that as quickly as titrating sturdy acids with sturdy bases the graph is going somewhat steep in the time of the equivalent ingredient. you will see this once you utilize a burret, like one drop and the indicator used will substitute, in our labs we usually phenol purple i've got self assurance and its like a faint purple shade your aiming for, the two way sturdy success.