what is the ph of 1.0 m acetic acid and 1.0 sodium acetate. (ka=1.8x10-5)
To find the pH of 1.0 M acetic acid and 1.0 M sodium acetate, we need to use the Henderson-Hasselbalch equation. The Henderson-Hasselbalch equation relates the pH of a solution to the concentration of the acid and its conjugate base.
The Henderson-Hasselbalch equation is given by:
pH = pKa + log([A-]/[HA])
Where:
pH = the pH of the solution
pKa = the logarithmic value of the acid dissociation constant (Ka)
[A-] = the concentration of the conjugate base (acetate, CH3COO-)
[HA] = the concentration of the acid (acetic acid, CH3COOH)
In this case, the Ka for acetic acid is given as 1.8 × 10^-5.
The first step is to calculate the ratio of [A-]/[HA]. Since we have 1.0 M acetic acid and 1.0 M sodium acetate, the concentration of the base ([A-]) and the acid ([HA]) are the same. Therefore, the ratio is 1.
Substituting the values into the Henderson-Hasselbalch equation, we get:
pH = pKa + log(1)
Since the logarithm of 1 is 0, the equation simplifies to:
pH = pKa
Therefore, the pH of the solution is equal to the pKa. In this case, the pKa is equal to the logarithm of the acid dissociation constant, which is -log(1.8 × 10^-5).
Calculating this value, we get:
pH = -log(1.8 × 10^-5) = approximately 4.74
So, the pH of both the 1.0 M acetic acid and 1.0 M sodium acetate solution is approximately 4.74.