Answer:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
Explanation:
The equilibrium of sodium acetate is:
CH₃COOH ⇄ CH₃COO⁻ + H⁺ Kₐ = 1,8x10⁻⁵
Where [CH₃COO⁻] are 0,1 M and 0,01 M and [H⁺] are 0,0025 M 0,005 M 0,01 M and 0,05 M.
For [CH₃COO⁻]=0,1 M and [H⁺]=0,0025M the concentrations in equilibrium are:
[CH₃COO⁻] = 0,1 M - x
[H⁺] = 0,0025 M - x
[CH₃COOH] = x
The expression for this equilibrium is:
Ka =
Replacing:
1,8x10⁻⁵ =
Thus:
0 = x²-0,102518x +2,5x10⁻⁴
Solving:
x = 0,100 ⇒ No physical sense
x = 0,0024995
Thus, [H⁺] = 0,0025-0,0024995 = 5x10⁻⁷
pH = - log [H⁺] = 6,30
Following the same procedure changing both [CH₃COO⁻] and [H⁺] initial concentrations the obtained pH's are:
[CH₃COO⁻] [H⁺] pH
0,1 M 0,0025 M 6,30
0,1 M 0,005 M 6,02
0,1 M 0,01 M 5,70
0,1 M 0,05 M 4,74
0,01 M 0,0025 M 5,22
0,01 M 0,005 M 4,75
0,01 M 0,01 M 3,38
0,01 M 0,05 M 1,40
I hope it helps!