Which of the following when mixed, will given a solution with `pH gt 7`.

Options

A. `0.1M HCI + 0.1M NaCI`
B. `100 mL of 0.1 M H_(2)SO_(4) + 100 mL of 0.3 M NaOH`
C. `100 mL of 0.1 M HC_(2)H_(3)O_(2) + 100 mL of 0.1 M KOH`
D. `25 mL of 0.1 HNO_(3) +25 mL of 0.1 M NH_(3)`

Answer

C

Explanation

a. It is a mixture of `S_(A) +` salt of `S_(A)//S_(B)`. So `pH` is due to `S_(A)` and is less than `7`.
b. `mEq of H_(2)SO_(4) = 100 xx 0.2 xx2 = 40`
`mEq of NaOH = 100 xx 0.3 = 30`
So `30 mEq of NaOH` will react with `30mEq` of `H_(2)SO_(4)`,
So `10 mEq` of `H_(2)SO_(4)` (strong acid will be left)
So `pH` will be less than `7`.
c. `mEq of CH_(3)COOH = 100 xx 0.1 = 10`
`mEq of KOH = 100 xx 0.1 = 10`
So `10 mEq of W_(A (CH_(3)COOH)` will react with `10 mEq` of `KOH (S_(B))` to give `10 mEq` of salt `(CH_(3)COONa)`. It is a salt of `W_(A)//W_(B)`.
So its `pH` will be greater than `7`.
`pH = (1)/(2) (pK_(w)+pK_(a) + logC)`
d. `mEq of HNO_(3) = 25 xx 0.1 = 2.5`
`mEq of NH_(3) = 25 xx 0.1 = 2.5`
So `2.5 mEq of HNO_(3)` will react with `2.5 mEq of NH_(3)` to given `2.5 mEq` of `NH_(4)NO_(3)`, which is a salt of `W_B)//S_(A)`, and its `ph` is always less than `7`.
`pH = (1)/(2) (pK_(w) - pK_(b) - logC)`