Question

It is desired that an acetic acid sodium acetate buffered solution have a pH of 5.27. You have a solution that has an acetic acid concentration of 0.01 M. What molarity of sodium acetate will you need to add to the solution, given that the pKa of acetic acid is 4.74

Answer 1

Answer:

0.034 M is the molarity of sodium acetate needed.

Explanation:

The pH of the buffer solution is calculated by the Henderson-Hasselbalch equation:  

$pH = pK_a + \log \frac{[A^-]}{[HA]}$

Where:  

pK_a= Negative logarithm of the dissociation constant of a weak acid  

$[Ac^-]$ = Concentration of the conjugate base  

[HA] = Concentration of the weak acid

According to the question:

$HAc(aq)\rightleftharpoons Ac^-(aq)+H^+(aq)$

The desired pH of the buffer solution = pH = 5.27

The pKa of acetic acid = 4.74

The molarity of acetic acid solution = [HAc] = 0.01 M

The molarity of acetate ion =$[Ac^-] = ?$

Using Henderson-Hasselbalch equation:  

$5.27= 4.74 + \log \frac{[Ac^-]}{[0.01 M]}$

$[Ac^-]=0.0339 M\approx 0.034M$

Sodium acetate dissociates into sodium ions and acetate ions when dissolved in water.

$NaAc(aq)\rightarrow Na^+(aq)+Ac^-(aq)$

$[Ac^-]=[Na^+]=[NaAc]= 0.034M$

0.034 M is the molarity of sodium acetate needed.