Calculate the pH of `0.08 M` solution of hypochlorous acid, HOCl. The ionization constant of the acid is `2.5 xx 10^(-5)`. Determine the percent dissociation of HOCl.

Explanation

Using direct formula,
`K_(a) = 2.5 xx 10^(-5) = 25 xx 10^(-6) = (5)^(2) xx 10^(-6)`
`pK_(a) =- log [(5)^(2) xx 10^(-6)] =- 2log 5+ 6`
`=- 2xx 0.7 +6 = 4.6`
a. `pH = (1)/(2) (pK_(a) - log C) = (1)/(2) (4.6 - log 0.08)`
`= (1)/(2) (4.6 - log (2)^(3) xx 10^(-2))`
`= (1)/(2) [4.6 - 3xx0.3 +2]`
`= 2.85`
`:. pH = 2.85`
b. `alpha = sqrt((K_(a))/(c)) = sqrt((25 xx 10^(-6))/(0.08)) = sqrt((25 xx10^(-6)xx100)/(8))`
`= sqrt(25xx10^(-4))/(2sqrt(2))`
`= (5xx10^(-2))/(2xx1.414) = 0.017`
`= 1.7 xx 10^(-3)M`
`HOCl(aq) + H_(2)O(l) hArr H_(3)O^(o+)(aq) + ClO^(ɵ)(aq)`
`{:("Initial conc",rArr0.08,-,0,0),("Conc at eq",rArr0.08(1-alpha),-,Calpha,Calpha),(,~~0.08(1-alpha~~1),,,):}`
`:.[HOCI] = 0.08M` ltbRgt `[H_(3)O^(o+)] = [CI^(Θ)] = Calpha = 0.08 xx 1.7 xx 10^(-3)`
`= 1.41 xx 10^(-3)M`
c. % dissociation `= ([HOCI]_("dissociated"))/([HOCI]_("undissociated")) xx 100`
`= (1.41 xx 10^(-3))/(0.08) xx 100 = 1.76%`