Question

The pH of a solution obtained by dissolving 0.1 mole of an acid HA is 100 ml of the aqueous solution was found to be 3.0 . Calculate the dissociation constant of the acid.

Answer 1

`pH = - log [H_(3)O^(+)] :. Log [H_(3)O^(+)]=-pH = - 3.0 ` [pH = 3.0 , given]
or `[H_(3)O^(+)]` = antilog (-3) = antilog `bar(3) = 10^(-3)` g ions/litre = 0.001 g ion/litre
Original conc.of the acid HA = 0.1 mole in 100 ml. = 1 mole/litre
`{:("HA dissociates as :",HA" "+" "H_(2)O,hArr,H_(3)O^(+)+,A^(-)),("Initial conc. :",1 M,,0,0),("Conc. at equilibrium:",1-0.001M,,0.001M,0.001M),(,,,,):}`
`:.` Dissociation constant (K) will be given by
`K=([H_(3)O^(+)][A^(-)])/([HA])=(0.001xx0.001)/(1-0.001)=(10^(-6))/(1)=10^(-6)` (Neglecting 0.001 in comparison to 1)