I need to find the pKa and Ka value of CH3COOH for my Acid-Base titration experi

Question

I need to find the pKa and Ka value of CH3COOH for my Acid-Base titration experiment. It’s pH before titration is 2.37 pH at equivalence point is 9.17 [H3O+] before titration is .00427 [H3O+] at equivalence point is 0.99 (Don’t know if it’s needed but the midpoint on the titration curve is (5.5ml, 7.49ph) . . Step by step explanation would be helpful!
I need to find the pKa and Ka value of CH3COOH for my Acid-Base titration experiment. It’s pH before titration is 2.37 pH at equivalence point is 9.17 [H3O+] before titration is .00427 [H3O+] at equivalence point is 0.99 (Don’t know if it’s needed but the midpoint on the titration curve is (5.5ml, 7.49ph) . . Step by step explanation would be helpful!
I need to find the pKa and Ka value of CH3COOH for my Acid-Base titration experiment. It’s pH before titration is 2.37 pH at equivalence point is 9.17 [H3O+] before titration is .00427 [H3O+] at equivalence point is 0.99 (Don’t know if it’s needed but the midpoint on the titration curve is (5.5ml, 7.49ph) . . Step by step explanation would be helpful!

Explanation / Answer

If your aim is to determine pKa and Ka , it is easy to determine

You know of equivalence point from graph

Half equivalence point = equivalence point/2

at half equivalence point

pH = pKa

reason for this is from Henderson - Hasselbalch equation

pH = pka + log([A-]/[HA])

[A-] =[ CH3COO- ]

[HA] =[ CH3COOH]

at half equivalence point

[CH3COO- ] = [ CH3COOH ]

so,

pH = pKa + log(1)

pH = pKa

So, you note down the pH corresponding to half equivalence point and that pH is the pKa value

pKa = - logKa

From this relation you can find out Ka value

Accepted value of pKa and Ka value of CH3COOH are

pKa = 4.75

Ka = 1.8×10^-5

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