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:
[tex]pH = pK_a + \log \frac{[A^-]}{[HA]}[/tex]
Where:
pK_a= Negative logarithm of the dissociation constant of a weak acid
[tex][Ac^-][/tex] = Concentration of the conjugate base
[HA] = Concentration of the weak acid
According to the question:
[tex]HAc(aq)\rightleftharpoons Ac^-(aq)+H^+(aq)[/tex]
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 =[tex][Ac^-] = ?[/tex]
Using Henderson-Hasselbalch equation:
[tex]5.27= 4.74 + \log \frac{[Ac^-]}{[0.01 M]}[/tex]
[tex][Ac^-]=0.0339 M\approx 0.034M[/tex]
Sodium acetate dissociates into sodium ions and acetate ions when dissolved in water.
[tex]NaAc(aq)\rightarrow Na^+(aq)+Ac^-(aq)\\[/tex]
[tex][Ac^-]=[Na^+]=[NaAc]= 0.034M[/tex]
0.034 M is the molarity of sodium acetate needed.
