1.

Calculate the degree of ionisation and `[H_(3)O^(+) ]` of 0.01 M acetic acid solution . `K_(a)` for acetic at 298 K is `1.8 xx 10^(-5)`

Answer» The ionisation of acetic acid is represented as :
`{:(,CH_(3)COOH(aq) ,hArr, CH_(3)COO^(-) (aq) ,+, H^(+) (aq)),("Initial molar concentration",0.01,,0,,0),("Equilibrium molar concentration",0.01(1-alpha),,0.01alpha,,0.01alpha):}`
Applying Law of chemical equilibrium,
`K_(a) = [[CH_(3)COO^(-)(aq)][H^(+)(aq)]]/[[CH_(3)COOH(aq)]] , 1.8 xx 10^(-5) = ((0.01alpha)(0.01alpha))/(0.01(1-alpha))`
Since `alpha` is very small for weak electrolyte `(CH_(3)COOH) ," therefore " (1-alpha) ~~ 1`.
`:. " "1.8 xx 10^(-5) = ((0.01alpha)(0.01alpha))/(0.01) = 0.01alpha^(2)`
`"or "" "alpha^(2) =(1.8xx 10^(-5))/(0.01) =1.8 xx 10^(-3) = 18 xx 10^(-3) = 18 xx 10^(-4)`
` " or "" "alpha sqrt(18xx10^(-4)) = 4.24 xx 10^(-2)`
`"or "" "[H_(3)O^(+)]=[H^(+)]Calpha = 0.01 xx 4.24 xx 10^(-2) = 4.24 xx 10^(-4)M`


Discussion

No Comment Found

Related InterviewSolutions