Calculate the pH of a given mixtures.
(a) `(4g CH_(3)COOH+6g CH_(3)COONa)` in 100 mL of mixture, (`K_(a)` for `CH_(3)COOH=1.8xx10^(-5))`
(b) 5 mL of `0.1MBOH+250` mL of `0.1 MBCI`, (`K_(a)` for `MOH=1.8xx10^(-5))`
(c ) (`0.25` mole of `CH_(3)COOH+0.35` mole of `CH_(3COOH=3.6xx10^(-4))`

Explanation

(a) We have pH= -log `K_(a)+log (["Salt"])/(["Acid"])`
`:. ["Salt"]= (6xx1000)/(82xx100)M` and `["Acid"]=(4xx1000)/(60xx100)M`
`:. pH= -log 1.8xx10^(5)+log(6xx1000//82xx100)/(4xx1000//60xx100)`
`:. pH= 4.7851`
(b) `pOH= -log K_(b)+log ((["Salt"])/(["Base"]))`
`.: "Total volume after mixing"=250+5=255mL`
Meq.of salt `= 250xx0.1=25`
Meq.of base `=5xx0.1=0.5`
`:. ["Salt"]= (25)/(255)`
`:. pOH= -log 1.8xx10^(-5)+log((25//255)/(0.5//255))`
pOH= 6.4437`
`:. pOH=14-pOH=7.5563`
(c ) `pH= -log K_(b)+log ((["Salt"])/(["Acid"]))`
`= -llog 3.6xx10^(-4)+log ((0.35//500)/(0.25//500))`
`pH= 3.5898`