The pH of 0.1 M solution of a weak acid `(HA)` is 4.50. It is neutralised with certain ammount of `NaOH` solution to decrease the acid content to half of initial value. Calculate the pH of the resulting solution.

Options

A. `4.50`
B. `8.00`
C. `7.00`
D. `10.00`

Answer

D

Explanation

`{:(HA,+,NaOH,hArr,NaA,+,H_(2)O,),("Weak acid",,"Strong base",,"Salt",,"Water","Final"),(0-1-0.05,,0.05-0.05,,0.05,,0.05,):}`
The resulting solution will be a buffer solution of weak acid and salt.
`pH = pK_(a) + log . (["Salt"])/(["acid"])`
`pH = pK_(a)`
Inital pH = 4.5
`[H^(+)] = 10^(-4.5) = alpha ` for weak acid
`K_(a) = C.alpha^(2) = 0.1 (10^(-4.5))^(2) = 10^(-10)`
`pH = pK_(a)= - log K_(a) = - log10^(-10) = 10 `