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Enough of a monoprotic acid is dissolved in water to produce a 0.0136 M solution. The pH of the resulting solution is 2.49. Calculate the Ka for the acid.

Respuesta :

Answer:

Ka=7.67E-4

Explanation:

monoprotic acid: HCl

⇒ HCl + H2O → H3O+ + Cl- + OH-

⇒ Ka = ([ H3O+] * [ Cl-] ) / [ HCl ].....(1)

∴ pH = 2.49 = -log [ H3O+ ] ⇒ [ H3O+ ] = 3.235E-3 M

load Balance: [ H3O+ ] = [ Cl- ] + [ OH- ]....[ OH- ] is despised as adding

⇒ [H3O+ ] = [ Cl- ]......(2)

replacing (2) en (1):

Ka = [ H3O+ ]² / [ HCl ] = (3.235E-3)² / (0.0136) = 7.67E-4

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