Derive the equation for calculation of pH of acidic buffer solution.

Explanation

Acetic acid and sodium acetate mixture act as acidic buffer solution. And its pH is approximate 4.75
The relation between pH, `K_a` , HA and `A^-` is given as follows :
The equilibrium of ionization of weak acid HA in water.
`HA+H_2O hArr H_3O^(+) + A^(-)`
`K_a=([H_3O^+][A^-])/([HA])`
`therefore [H_3O^+]=K_a ([HA])/([A^-])`
taking log both the side
log `[H_3O^+]=log K_a + "log" "[HA]"/([A^-])`
taking minus sign both the side,
`-log [H_3O^+]=-log K_a-log ([HA])/([A^-])`
`therefore pH=pK_a + "log" ([A^-])/([HA])`....(Eq.-i)
`pH=pK_a + "log" "[Conjugate base]"/"[Acid [HA]]"`....(Eq.-ii)
This equation -(ii) is called Hendrson-Hasselbalch equation
In this equation of value :
(i)`([A^-])/([HA])` is the ratio of concentration of conjugate base concentration of weak acid (negative ion).
(ii)[HA] is weak acid . So, it undergo very less ionization so the concentration of [HA] is negligible compare to initial concentration .
(iii) The around of conjugate base is due to ionization of salt . So the concentration of conjugate base is also differ from of initial concentration . So equation can be ,
pH=`pK_a+"log" "[Salt]"/"[Acid]"`.....(Eq.-iii)