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

`pOH =-log K_(b)+log (["salt"])/(["base"])`
Total volume after mixing =250+5=255mL
Meq. Of salt `=250 xx 0.1=25`
Meq. Of base `=5 xx 0.1 =0.5`
`["salt"]=(25)/(255) and ["base"]=(0.5)/(255)`
`therefore pOH =-log 1.8 xx 10^(-5) +log (25//255)/(0.5//255)`
pOH =6.4437
`therefore pOH =14-pOH=7.5563`