Calculate the `pH` of the following mixtures given `K_(a) = 1.8 xx 10^(-5)` and `K_(b) = 1.8 xx 10^(-5) (pK_(a) = pK_(b) = 4.7447)`.
a. `50mL 0.05 M NaOH +50mL` of `0.1 M CH_(3)COOH`
b. `50mL 0.1 M NH_(4)OH +50 mL` of `0.05 MHCI`

Explanation

`{:(,CH_(3)COOH,+,NaOH,to,CH_(3)COoNa,+,H_(2)O),("Intial",50 xx 0.01 ,,50 xx 0.1 ,,0,,0),("Millimoles" ,=5,,=2.5,,0,,0),("Millimotes ",,,,,,,),("after reaction ",2.5,,0,,2.5,,2.5):}`
The solution consists of `CH_(3)COOH` and `CH_(3)COONa` and thus for buffer
`pH = pK_(a) +log. (["Salt"])/(["Acid"]) `
` pH = pK_(a) + log. ([CH_(3)COONa])/([CH_(3)COOH])`
` rArr pH = -log 1.8 xx 10^(-5) + log. (2.5 //100)/(2.5//100)`
` rArr pH = -log log 1.8 xx10^(-5) + log . (2.5//100)/(2.5//100)`
`rArr pH = 4.7447`
(b) `{:(,CH_(3)COOH,+,NaOH,to,CH_(3)COONa,+,H_(2)O),("Intial millimoles ",50xx0.1,,50 xx 0.1,,0,,0),(,=5,,=5,,0,,0),("Final millimoles ",0,,0,,5,,5):}`
Finally concentration of `CH_(3)COONa = 5/100` and pH is decided by salt hydrolysis .
`{:(CH_(3)COONa,+,H_(2)O,hArr,CH_(3)COOH,+,NaOH),(C,,,,C,,0),(C(1-h),,,,Ch,,Ch):}`
`[OH^(-)] = Ch = C sqrt((K_(h))/C) = sqrt((K_(w)xxC)/(K_(a)))= sqrt((10^(14))/(1.8 xx10^(-5))xx5/100) = 5.27 xx10^(-6) M `
`[H^(+)] = (10^(-14))/(5.2 xx 10^(-5)) = (10^(-8))/ (5.2)`
` :. pH = 8.7218 `