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.

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

it's filtration, but this is an assumption so my apologies if i got this