Question

Buffer capacity. Two solutions of sodium acetate are prepared, one having a concentration of 0.1 M and the other having a concentration of 0.01 M. Calculate the pH values when the following concentrations of HCl have been added to each of these solutions: 0.0025 M, 0.005 M, 0.01 M, and 0.05 M.

Answer 1

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 = $\frac{[CH3COO^-] [H^+] }{[CH3COOH]}$

Replacing:

1,8x10⁻⁵ = $\frac{[0,1-x] [0,0025-x] }{[x]}$

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!