Calculate `pH` of
a. `0.002 N CH_(3) COOH` having `2.3%` dissociation.
b. `0.002N NH_(4) OH` having `2.3%` dissociation.

Explanation

a. `0.002 N CH_(3)COOH`: Acetic acid is weak electrlyte and partially associated.
`{:(,CH_(3)COOHhArr,CH_(3)COO^(Theta)+,H^(o+),),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}`
`:. [H^(o+)] = C alpha = 2 xx 10^(-3) xx (2.3)/(100) = 4.6 xx 10^(-5) M`
`:. pH = -log[H^(oplus)] = - log(4.6 xx 10^(5))`
`pH = 4.3372`
b. `0.002 N NH_(4)OH: NH_(4)OH` is weak base and partially dissociated.
`{:(,NH_(4)OHhArr,NH_(4)^(Theta)+,overset(Theta)OH,),("Conc before dissocitation",1,0,0,),("Conc after dissociation",1-alpha,alpha,alpha,):}`
`:. [overset(Theta)OH] = C alpha = 2 xx 10^(-3) xx (23)/(100) = 4.6 xx 10^(-5) M`
`because pOH = log [overset(Theta)OH] =- log (4.6 xx 10^(-5))`
`:. pOH =- 4.3372 :. pH = 14 - 4.3372`
`pH = 9.6628`