Question

What is the ratio of acetic acid to acetate ion in solutions at pH 4, 6, 9 if the pka of acetate is 4.8?

Answer 1

At pH = 4, [Acetic acid]/[Acetate ion] = 6.3096

At pH = 6, [Acetic acid]/[Acetate ion] = 0.0631

At pH = 9, [Acetic acid]/[Acetate ion] = 6.3096×10⁻⁵

Step-by-step explanation:

The pH of a buffer solution is calculated using Henderson Hasselbalch equation. The expression for the equation is:

`pH=pK_a+log([Concentration\ of\ conjugate\ base])/([Concentration\ of\ acid])`

For, Acetic acid (pKa = 4.8), Acetate will be its conjugate base.

To calculate the ratio of acetic acid and acetate ion in the solution.

Using the property of log that:

log(a/b) = -log(b/a)

The equation can be written as:

`pH=pK_a-log([CH_3COOH])/([CH_3COO^-])`

For pH =4 :

`4=4.8-log([CH_3COOH])/([CH_3COO^-])`

`log([CH_3COOH])/([CH_3COO^-])=0.8`

Thus, Taking anti log and solving as:

`([CH_3COOH])/([CH_3COO^-])=6.3096`

[Acetic acid]/[Acetate ion] = 6.3096

For pH =6 :

`6=4.8-log([CH_3COOH])/([CH_3COO^-])`

`log([CH_3COOH])/([CH_3COO^-])=-1.2`

Thus, Taking anti log and solving as:

`([CH_3COOH])/([CH_3COO^-])= 0.0631`

[Acetic acid]/[Acetate ion] = 0.0631

For pH =9 :

`9=4.8-log([CH_3COOH])/([CH_3COO^-])`

`log([CH_3COOH])/([CH_3COO^-])=-4.2`

Thus, Taking anti log and solving as:

`([CH_3COOH])/([CH_3COO^-])= 6.3096* 10^(-5)`

[Acetic acid]/[Acetate ion] = 6.3096×10⁻⁵