Percentange ionisation of weak acid can be calculated using the formula:

Options

A. `100sqrt((K_(a))/(C))`
B. `(100)/(1+110^((pK_(a)-pH)))`
C. Both (A ) and (B)
D. None of these

Answer

C

Explanation

`{:(,HA,hArr,H^(+),A^(-)),("Initial",C,,,),("At eqm. ",C(1-alpha),,Calpha,Calpha):}`
`K_(a)=([H^(+)][A^(-)])/([HA])~~Calpha^(2)`
`[1-alpha~~1]`
`alpha=sqrt((K_(a))/(C))` and `C=(K_(a))/(alpha^(2))`
`% alpha =100 sqrt((K_(a))/(C ))`
If `alph` is not very small
`K_(a)=(Calpha^(2))/(1-alpha)`
`K_(a)=([H^(+)]alpha)/(1-alpha)`
`(1-alpha)/(alpha)=(H^(+))/(K_(a))`
`(1)/(alpha)-1=(H^(+))/(K_(a))`
`(1)/(alpha)=(H^(+))/(K_(a))+1`
`(1)/(alpha)=(10^(-pH))/(10^(-pK_(a)))+1`
`(1)/(alpha)=10^(pK_(a)-pH)+1=(1)/(10^(pK_(a)-pH)+1)`
`:. % alpha=(100)/(10^(pK_(a)-pH)+1)`