pH of 0.2 M acetic acid solution will be (pK, of CH3COOH = 4.74)
Answers
Answered by
0
Answer:
CH
3
COOH+NaOH→CH
3
COONa+H
2
O
Moles of CH
3
COOH=0.2×100×1=20 mmol
Moles of NaOH=0.2×100×1=20 mmol
So there is complete neutralisation
[salt]=C=
100+100
20
=
200
20
=0.1M
pH=7+
2
1
pK
a
+
2
1
log
10
C
=7+
2
1
(4.74)+
2
1
log(0.1)
=7+2.37−
2
1
log
10
10
pH=9.37−0.5=8.87
Explanation:
Similar questions