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

Answer:

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

Explanation:

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

$pH=pK_a+log\frac{[Concentration\ of\ conjugate\ base]}{[Concentration\ of\ acid]}$

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:

$pH=pK_a-log\frac{[CH_3COOH]}{[CH_3COO^-]}$

For pH =4 :

$4=4.8-log\frac{[CH_3COOH]}{[CH_3COO^-]}$

$log\frac{[CH_3COOH]}{[CH_3COO^-]}=0.8$

Thus, Taking anti log and solving as:

$\frac{[CH_3COOH]}{[CH_3COO^-]}=6.3096$

[Acetic acid]/[Acetate ion] = 6.3096

For pH =6 :

$6=4.8-log\frac{[CH_3COOH]}{[CH_3COO^-]}$

$log\frac{[CH_3COOH]}{[CH_3COO^-]}=-1.2$

Thus, Taking anti log and solving as:

$\frac{[CH_3COOH]}{[CH_3COO^-]}= 0.0631$

[Acetic acid]/[Acetate ion] = 0.0631

For pH =9 :

$9=4.8-log\frac{[CH_3COOH]}{[CH_3COO^-]}$

$log\frac{[CH_3COOH]}{[CH_3COO^-]}=-4.2$

Thus, Taking anti log and solving as:

$\frac{[CH_3COOH]}{[CH_3COO^-]}= 6.3096\times 10^{-5}$

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