a. i. When no titration is carried out (it is a case of `W_(A))`
`:. pH_(W_(A)) = (1)/(2) (pK_(a) - logC)`
`= (1)/(2) (4.74 - log.(1)/(10))`
`= (1)/(2) (4.74 - log 10^(-1)) = (1)/(2) (4.74 +1)`
`= 2.87`
ii. When `16mL` of `NaOH` is added (acidic buffer is formed)
mmol of `CH_(3)COOH = 20 xx (1)/(10) = 2`
mmol of `NaOH = 16 xx (1)/(10) = 1.6`
`1.6` mmol of `NaOH` react with `1.6` mmol of `Ch_(3)COOH`, so 6 mmol of `CH_(3)COONa` is formed and `(2-1.6) = 0.4 "mmol"` of `CH_(3)COOH (W_(A))` si left in the solution.
`:. pH_("acidic buffer") = pK_(a) + log. (["Salt"])/(["Acid"])`
[Total `V = 20 +16 = 36 mL`]
`= 4.74 + log .[(1.6//36)/(0.4//36)]`
`= 4.74 + log 4`
`= 4.74 + 2 xx 0.3 = 5.34`
ii. When `20mL` of `NaOh` is added (salt of `W_(A)//S_(B)` is formed)
mmol of `CH_(3)COOH = 20 xx (1)/(10) = 2`
mmol of `NaOH = 20 xx (1)/(10) = 2`
`["Salt"] = [CH_(3)COONa] = (2mL)/((20+20)mL)`
`= (2)/(40) = 0.05M`
`pH = (1)/(2) (pK_(w) +pK_(a) + logC)`
`pH = (1)/(2) (14 + 4.74 + log (0.05)]`
`= (1)/(2) (18.74 + log 5 xx 10^(-2))`
`(log 5 xx 10^(-2) = 2 - 0.7 = 1.3)`
`= (1)/(2) (18.74 +1.3) = 10.02`
iv. When `30mL` of `NaOH` is added (it is a case of `S_(B)`)
mmol of `CH_(3)COOH = 20 xx (1)/(10) = 2`
mmol of `NaOH = 30 xx (1)/(10) = 3`
mmol of `NaoH` left `= 3 - 2 = 1`
`[NaOH] = [overset(Θ)OH] = (1)/(50) = 0.02 M = 2 xx 10^(-2)M`
`pOH =- log (2xx10^(-2)) = 1.7`
`pH = 14 - 1.7 = 12.3`.
b. i. No titration is carried out (it is a case of `S_(B)`)
`[NaOH] = (1)/(10) = 10^(-1)`
`pOH = 1, pH = 14 - 1 = 13`
ii. When `18mL` of `CH_(3)COOH` is added (it is case of `S_(B)`)
mmol of `NaOH = 20 xx (1)/(10) = 2`
mmol of `CH_(3)COOH = 18 xx (1)/(10 = 1.8`
mmol of `NaOH` left `= 2 - 1.8 = 0.2`
`[overset(Θ)OH] = (0.2 "mmol")/((20+18) ml) = 0.005 = 5 xx 10^(-3)`
`pOH =- log (5 xx 10^(-3)) = 3 - 0.7 = 2.3`
`pH = 14 - 2.3 = 11.7`
iii. When `20mL` of `CH_(3)COOH` is added (salt of `W_(A)//S_(B)` is formed)
Same answer as in part (a), case (iii)
`pH = 10.02`.
iv. When `40mL` of `CH_(3)COOH` is added (acidic buffer is formed)
mmol of `NaOH = 2`
mmol of `CH_(3)COOH = 40 xx (1)/(10 = 4`
mmol of `Ch_(3)COOH` left `= 4 - 2 = 2`
mmol of salt `(CH_(3)COONa)` formed `= 2`
`pH = 4.74 + log ((2//60)/(2//60))`
`pH = 4.74`
c. i. When no titration is carried out (it is case of `W_(B)`)
`[NH_(4)OH] = (1)/(10) = 10^(-1)M`
`pOH_(W_(B)) = (1)/(2)(pK_(b) - log C) = (1)/(2) (4.76 - log 10^(-1))`
`= (1)/(2) (4.76 +1) = 2.88`
`pH = 14 - 2.88 = 11.12`
ii. When `4mL` of `H_(2)SO_(4)` is added (basic buffer is formed)
mmol of `NH_(4)OH = 10 xx (1)/(10) = 1`
mmol of `H_(2)SO_(4) = 4 xx (1)/(10) = 0.4`
`{:(,2NH_(4)OH+,H_(2)So_(4)rarr,(NH_(4))_(2)SO_(4)+,2H_(2)O),("Final",1mmol,0.4mmol,0,-),("Final",(1-2xx0.4),(0.4-0.4),0.4mmol,-),(,=0.2,=0,,):}`
`:. [NH_(4)OH]_("left") = 0.2` mmol
`[(NH_(4))_(2)SO_(4)] = 0.4` mmol
`[overset(o+)NH_(4)] = 2 xx 0.8` mmol
Total `V = 10 +4 = 14mL`
`pOH_("basic buffer") = pK_(b) + "log" [(0.8//14)/(0.2//14)]`
`= 4.76 + log 4`
`= 4.76 +2 xx 0.3`
`= 5.36`
`pH = 14 - 5.36 = 8.64`
Second method : mmoles `= mEq` of `NH_(4)OH = 1`.
`mEq` of `H_(2)SO_(4) = 4 xx (1)/(10) xx2 ("n factor of" H_(2)SO_(4))`
`= 0.8`
mmol = mEq of `[NH_(4)OH]_("left") = 1- 0.8 = 0.2`
It is evident that `0.8 mEq` of `H_(2)So_(4)` reacts with `0.8mEq` of `NH_(4)OH (~~0.8 mmol` of `NH_(4)OH` since 'n' factor is one) to given `0.8 mEq` of `(NH_(4))_(2)SO_(4)`.
`pOH_("basic buffer") = pK_(b) + "log" ([0.8//14)/(0.2//14)] = 5.36`
`pH = 14 - 5.36 = 8.64`
iii. When `5mL` of `H_(2)SO_(4)` is added (salt of `W_(B)//S_(A)` is formed)
According to equaitons :
`{:(,2NH_(4)OH+,H_(2)SO_(4)rarr,(NH_(4))_(2)SO_(4)+,2H_(2)O),("Final",1mmol,5xx(1)/(10),0,0),("Final",(1-2xx0.5),0.5-0.5,0.5mmol,),(,=0,=0,,):}`
`[(NH_(4))_(2)SO_(4)] = 0.5 mmol`,
`[overset(o+)NH_(4)] = 2 xx 0.5 = 1mmol`
`:. [overset(o+)NH_(4)] = (1 "mmol")/((10+5)mL) = 0.06M`
Alternatively :
mEq = mmol of `NH_(4)OH = 1`
mEq of `H_(2)SO_(4) = 5 xx (1)/(10) xx2 ("n factor of "H_(2)SO_(4)) =1`
`["Salt"] = [(NH_(4))_(2)SO_(4)]`
`= [overset(o+)NH_(4)] = (1mEq)/(15mL) = 0.06M`
`pH = (1)/(2) (pK_(w) - pK_(b) - logC)`
`= (1)/(2) (14 - 4.76 - log (6 xx 10^(-2))`
`= (1)/(2) (14 - 4.76 +1.22) = 5.23`
`[{:(-(log6xx10^(-2)),=-log3-log2+2,,,),(,=-0.48-0.3+2,,,),(,=1.22,,,):}]`
iv. When `10mL` of `H_(2)SO_(4)` is added (it is a case of `S_(A))`
mEq = mmoles of `NH_(4)OH = 1`
mEq of `H_(2)SO_(4) = 10 xx (1)/(10) xx 2 =2`
mEq of `H_(2)SO_(4)` left `= 2 - 1 = 1`
Total `V = 10 + 10 = 20 mL`
`[H^(o+)] = (1 mEq)/(20 mL) = 0.05 = 5 xx 10^(-2)`
`[pH] =- log (5xx10^(-2)) = 1.3`
d. i. When no titration is carried out (it is a case of `S_(A))`
`[H^(o+)] = (1)/(10) xx2 ("n factor of" H_(2)SO_(4))`
`= 2xx 10^(-1)`
`pH =- log (2 xx 10^(-1)) = 0.7`
ii. When `10mL` of `NH_(4)OH = 10 xx (1)/(10) = 1`
mEq of `H_(2)SO_(4)` left `= 2 - 1 = 1`
Total `V = 10 +10 = 20 mL`
`[H^(o+)] = (1mEq)/(20mL) = 0.05 = 5 xx 10^(-2)`
`pH =- log (5 xx 10^(-2)) = 1.3`
iii. When `20mL` of `NH_(4)OH` is added (salt of `W_(B)//S_(A)` is formed)
`mEq` of `H_(2)So_(4) = 2`
mEq = mmol of `NH_(4)OH = 20 xx (1)/(10) = 2`
`["Salt"] = [(NH_(4))_(2)SO_(4)] = (2mEq)/((10+20)mL) = 0.06N`
Same `pH = 5.23` (as in part C), (case iii)
iv. When `40mL` of `NH_(4)OH` is added (basic buffer is formed)
`mEq` of `H_(2)SO_(4) = 2`
mEq = mmol of `NH_(4)OH = 40 xx (1)/(10) = 4`
mEq of `NH_(4)OH` left `= 4 - 2 = 2`
`2mEQ` or `mmo` of `NH_(4)OH` reacts with `2mEq` of `H_(2)So_(4)` to given `2mEq` of `(NH_(4))_(2)SO_(4)`.
`pOH_("basic buffer") = pK_(b) + "log"[(2//50)/(2//50)]`
`pOH = pK_(b) = 4.76`
`pH = 14 - 4.76 9.24`
