I received a hint that I have to find pKb(Kb) and solve for the titration equiva

Question

I received a hint that I have to find pKb(Kb) and solve for the titration equivalence to then find Ka. And that it also has something to do with the titration break at 17 and 18mL. I'm so confused on how to do this.

Adobe Reader File Edit View Window Help chem3.pdf - + | 100% P Tools Q) The following data were obtained from the titration of a 25.0 mL week base with 0.1 M HCI Calculate Ka for the conjugate acid of this week base Volume of HCI pH (mL) 0 11.05 10.86 10.78 10.42 10.12 10.0 9.70 9.60 9.30 9.10 8.80 8.5 8.43 7.96 7.56 7.12 6.89 6.65 3.45 3.33 3.21 3.15 3.05 3.01 2.96 2.9 2 3 4 5 6 8 10 12 13 14 15 16 17 18 19 20 21 23 24 25

Explanation / Answer

Solution :-

At the volume of 17 and 18 ml of the HCl the pH chages drastically means the equivalence point of the titration is 18 ml of the HCl

So at the equivalence point all the base is converted into the conjugate acid

since equivalence point is at 18 ml then half equivalence point of the titration is reached at 18/2 = 9 ml

where the conjugate acid and base concentration is same

so according to the henderson equation the pH = pka at the half equivalence point

so the pH = 9.10 at the 9 ml volume

so pka = 9.10

now lets find the ka from the pka

pka = -log ka

ka= antilog [-pka]

ka= antilog [-9.10]

9.74*10^-10 = ka

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