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Calculate the `pH` of `0.08` solution of `HOCI` (hydrochlorous acid). The ionisation constant of the acid is `2.5 xx 10^(-5)`. Determine the percent dissociation of `HOCI`. |
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Answer» 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%` |
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