A modification on a chemical compound results in the shifting of its pKa value f

Question

A modification on a chemical compound results in the shifting of its pKa value from 2.0 to 4.0. Compared to the half-equivalence point for the original compound, the new compound will reach its half equivalence point when the hydronium ion concentration is:

A) 4x less than the original compound

B) 2x less than the original compound

C) 100x less than the original compound

D) 10x less than the original compound

Please explain and show work. Could I use the Henderson-Hasselbach equation to solve this?

Thanks!

Explanation / Answer

let the acid be HA

now, HA + H2O <======> (A-) + (H3O+)

Ka = [A-]*[H3O+]/[HA]

pKa = -logKa

thus, greater the value of pKa lesser will be the [H3O+]

thus, the correct option is :-

C) 100x less than the original compound

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