Calculate the `pH` of the following mixtures given `(pK_(a) = pK_(b) = 4.7447)`:
a. `50mL 0.1 M NaOH +50mL 0.1MCH_(3)COOH`
b. `50mL 0.1 m NaOH +50mL 0.05 M CH_(3)COOH`
c. `50mL 0.05M NaOH +50mL 0.1 M CH_(3)COOH`
d. `50mL 0.1M NH_(4)OH +50mL 0.05 MHCI`
e. `50mL 0.05M NH_(4)OH +50mL 0.1 MHCI`
f. `50mL 0.05 M NH_(4)OH + 50mL 0.05 M CH_(3)COOH`

Explanation

a. 5mEq of NaoH combines with `5mEq` of `CH_(3)COOH` and froms 5 mEq of salt of `W_(A)//S_(B)`.
`["salt"] = (5)/(100) = 0.05M`
`pH = (1)/(2) (pK_(w) +pK_(a) +logC)`
`= (1)/(2) (14 +4.7447 + log 0.05) = 87.21`
b. `NaOH = 50 xx0.1 = 5 mEq`
`CH_(3)COOH = 50 xx 0.05 = 2.5 mEq`
`2.5 mEq` of `CH_(3)COOH` reacts with 2.5 mEq of NaOH and 2.5 mEq of NaOH is left.
`:. [NaoH] = [OH^(Θ)] = (2.5)/(100) = 2.5 xx10^(-2)`
`pOH=- log (2.5 xx 10^(-2)) = 1.6021`
`pH = 14 - 1.6021 = 12.3979`
c. 2.5mEq of NaOH combines with 5mEq of `CH_(3)COOH` and forms 2.5 mEq of `CH_(3)COONa`
`:. 2.5 mEq` of `CH_(3)COOH` is left. so it forms acidic buffer.
`pH = pK_(a) + "log"[("Salt")/("Acid")]`
`= 4.7447 + log ((2.5)/(2.5)) = 47.447`
d. `NH_(4)OH = 50 xx0.1 = 5 mEQ`
`HCI = 50 xx 0.05 = 2.5 mEq`
`2.5 mEq` of `HCI` conbines with `2.5 mEq` of `NH_(4)OH` and forms a salt `NH_(4)CI` (salt of `S_(A)//W_(B)`) and `2.5 mEq` of `W_(B)(NH_(4)OH)` is left. So, it forms basic buffer.
`:. pOH = pK_(b) + log [("Salt")/("Base")]`
`= 4.7447 + log ((2.5)/(2.5)) = 4.7447`
`pH = 14 - 4.7447 = 9.2553`
e. `NH_(4)OH = 50 xx 0.05 = 2.5 mEq`
`HCI = 50 xx 0.1 = 5mEq`
`2.5 mEq` of `NH_(4)OH` reacts with `2.6 mEq` of `HCI` and
`:. [HCI] = [H_(3)O^(o+)] = (2.5)/(100) = 2.5 xx 10^(-2)`
`pH =- log (2.5 xx 10^(-2)) = 1.6021`
f. `NH_(4)OH = 50 xx 0.05 = 2.5 mEq`
`CH_(3)COOH = 50 xx 0.05 = 2.5 mEq`
`2.5 mEq` of `NH_(4)OH` reacts with `2.5mEq` of `CH_(3)COOH` and gives `2.5mEq` of salt `(CH_(3)COONH_(4)) ("salt of" W_(A))/ W_(B))` which hydrolyses.
`pH = (1)/(2) (pK_(w) + pK_(a) - pK_(b))`
`= (1)/(2) (14 + 4.7447 - 4.7447) = 7`