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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. |
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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 % |
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