1.

Weak Acid HA (K_a = 1.4 xx 10^(-5)) 0.1 M so in dissow in 2 lit. Find percentage of ionization and pH of solution.

Answer»

SOLUTION :`HA_((aq)) + H_2O_((l)) HARR H_3O_((aq))^(+) + A_((aq))^(-)`
`K_a=1.4xx10^(-5)` initial concentration C=0.1 M
`[H_3O^+]=SQRT(K_axxC)`
`=(1.4xx10^(-5)xx0.1)^(1/2)`
`=1.183xx10^(-3)` M
`pH=-log [H_3O^+]`
`=-log (1.183xx10^(-3))`
=-(0.07306-3)
=2.9269 `APPROX` 2.93
`K_a=(alpha^2C)/(1-alpha)` Now, `1-alpha approx 1`
`K_a=alpha^2C`
`alpha=sqrt(K_a/C) = sqrt((1.4xx10^(-5))/0.1)`
`alpha=1.183xx10^(-2)`
% of dissociation = `alpha xx 100`
`=1.183xx10^(-2) xx100`
=1.183 %


Discussion

No Comment Found