What is the pH of a solution that is 0.150 M in acetic solution and 0.300 M in sodium acetate? The Ka of acetic acid is 1.8 x 10^-5.
My work:
pH= pKa + log (base/acid)
pH= -log (1.8 x 10^-5) + log (0.300/0.150)
pH=5
I would not throw away the places after 5 that are eligible to be retained. I came up with 5.04 to three significant figures.
To calculate the pH of the solution, you can use the Henderson-Hasselbalch equation:
pH = pKa + log(base/acid)
Where:
pH is the measure of acidity or alkalinity on a logarithmic scale.
pKa is the negative logarithm (base 10) of the acid dissociation constant (Ka). It represents the strength of an acid.
base is the concentration of the base (sodium acetate in this case).
acid is the concentration of the acid (acetic acid in this case).
From the information provided, you know that the Ka of acetic acid is 1.8 x 10^-5. To find the pKa, take the negative logarithm of the Ka:
pKa = -log(1.8 x 10^-5)
pKa ≈ 4.74
Next, substitute the values into the Henderson-Hasselbalch equation:
pH = 4.74 + log(0.300/0.150)
pH = 4.74 + log(2)
pH ≈ 4.74 + 0.301
pH ≈ 5.04
Therefore, the pH of the solution is approximately 5.04.