If a solution were made by mixing equal volume of 0.1M acetic acid(HA) and 0.1M Na-acetate(A^-) what pH would it gives?
wouldn't they neutralize each other exactly?
To determine the pH of the solution made by mixing equal volumes of 0.1M acetic acid (HA) and 0.1M sodium acetate (A^-), we need to consider the nature of the acetic acid and its conjugate base, the acetate ion.
Acetic acid is a weak acid and partially dissociates in water, producing hydrogen ions (H+) and acetate ions (A^-). This can be represented by the equation:
HA ⇌ H+ + A^-
To find the pH of the solution, we have to consider the ionization constant of acetic acid, known as the dissociation constant (Ka). The Ka for acetic acid is approximately 1.8 x 10^-5.
Using the Henderson-Hasselbalch equation, we can calculate the pH of the solution:
pH = pKa + log ([A^-]/[HA])
In this case, since equal volumes of HA and A^- are mixed, the ratio [A^-]/[HA] = 1.
Now, to calculate the pKa, we can use the equilibrium expression for the dissociation of acetic acid:
Ka = [H+] [A^-] / [HA]
Since we are mixing equal volumes of 0.1M acetic acid and 0.1M sodium acetate, the concentrations of HA and A^- will be equal, which means [A^-] = [HA].
Thus, the equation can be simplified to:
Ka = [H+] [A^-] / [HA] = [H+] [HA] / [HA] = [H+]
Simplifying further, we find:
Ka = [H+], and taking the negative logarithm of both sides:
-log(Ka) = -log([H+])
Which gives us:
pKa = pH
Substituting this result into the Henderson-Hasselbalch equation, we can simplify it to:
pH = pKa + log(1) = pKa
Therefore, the pH of the solution made by mixing equal volumes of 0.1M acetic acid and 0.1M sodium acetate will be equal to the pKa of acetic acid. The pKa of acetic acid is approximately 4.76.
So, the pH of the solution will be approximately 4.76.