The `pH` of `0.1M` monobasic acid is `4.50`. Calculate the concentration of species `H^(+)`.
`A^(-)` and `HA` at equilibrium . Also, determine the value of `K_(a)` and `pK_(a)` of the monobasic acid.

Explanation

`pH=-log[H^(+)]`
Therefore, `[H^(+)]=10^(-pH)=10^(-4.50)`
`=3.16xx10^(-5)`
`[H^(+)]=[A^(-)]=3.16xx10^(-5)`
Thus, `K_(a)=[H^(+)][A^(-)]//[HA]`
`[HA]_("eqlbm")=0.1-(3.16xx10^(-5))-~=0.1`
`K_(a)=(3.16xx10^(-5))^(2)//0.1=1.0xx10^(-8)`
`pK_(a)=-log(10^(-8))=8`
Alternatively, "Percent dissociation" is another useful method for measure of strength of a weak acid and is given as :
Percent dissociation
`=[HA]_("dissociated")//[HA]_("initial")xx100%`