In which of the following mixed aqueous solutions pH = `pk_a` at equilibrium?
(1) `100 mL of 0.1 M CH_3 COOH + 100 mL `of` 0.1 M CH_3 COONa`
` (2) 100 mL o f 0.1 M CH_3 COOH + 50 mL o f 0.1 M NaOH `
(3) `100 mL o f 0.1 M CH_3 COOH + 100 mL o f 0.1 M NaOH `
` (4) 100 mL o f 0.1 M CH_3 COOH + 100 mL o f 0,1 M NH_3`

Options

A. (1) is correct.
B. (2) is correct.
C. (3) is correct.
D. Both (1) and (2) are correct.

Answer

A, B, D

Explanation

Let us analyse each case :
` (1) 100 m L ` of 0.01 `M CH_3 COOH + 100 m L ` of ` 0.1 M CH_3 COONa`
` [CH_3 COOH]=(100 xx 0.1)`
`[CH_3 COONa]=( 100 xx 0.1)`
the mixture is an acidic buffer
using henderson equation we get
` pH =pK_a + log ([CH_3 COONa])/([CH_3 COOH])`
` implies PH =pK_a + og ((100 xx 0.1 ))/( ( 100 xx 0.1 )) implies pH =pK_a + log 1`
`implies pH = pk_a ( :. log 1=0)`
(2) 100 mL of `0.1 M CH)3 COOH + 50 mL ` of `0.1 M NaOh `

Remaining
5 milliole 0 millimole 5 millimole 5 millimole
so an acidic buffer is forward
` therefore pH = pK_a + log (["salt"])/(["acid"])`
`implies pH =pK_a + log (5)/(5) implies PH = pK_a `
`( :. log (5)/(5) =- log 1=0)`
`(3) 100 m L ` of ` 0.1 M CH_3 COOH +100 mL ` of ` 0.1 M`
` NaOH `

Now `CH_3 COONa ` which is a salt from weak acid and strong base will undergo hydrolysis (anionic hydrolysis and the pH will be deter mined as below
`pH =7 +1/2 pK_a +1/2 log c`
so` pH ne pK_a `
(4) `100 mL ` of `0.1 M CH_3 COOH + 100 mL `of 0.1 M `NH_3`
`[CH_3] or [NH_4OH ] =( 100 xx 0.1 ) =10`
` (##MTG_WB_JEE_CHE_C12_E02_010_S03.png" width="80%">
Now `CH_3 COONa` which is a salt from weak acid and strong base will underogo hydrolysis ( anionic hydrolysis ) and the pH will be deteremined as below
`pH = 7 + 1/2 pK_a +1/2 log c`
So `PH ne pK_a `
(4) 100 mL of 0.1 `CH_3 COOH + 100 mL ` of `0.1 MCH_3`
`[CH_3 COOH ]=(100 xx 0.1 )=10`
`[NH_3 COOH ]=(100 xx 0.1 )=10`

`pH =7 +1/2 (pk_a-pK_b)`
so `pH ne pK_a `