A solution is prepared by dissolving 0.1 mol of ethanoic acid (CH3COOH) in 500ml of water at 25 C. pKa of ethanoic acid is 4.7.
What is the pH of the solution?
I'm assuming the equation is pH = pKa + log[Base]/[Acid] but can't get it to work?
Nope. The HH equation is for buffers. You have an acid listed but no base to use. This is just the ionization of an acid. CH3COONa = HAc
.........HAc ==> H^+ + Ac^-
I........0.2.....0......0
C.......-x.......x......x
E.......0.2-x....x......x
Substitute the E line into the Ka expression for Ka and solve for x, then convert to pH.
You are on the right track with using the Henderson-Hasselbalch equation to solve for the pH of the solution. The Henderson-Hasselbalch equation relates the pH of a solution to the pKa of an acid and the ratio of its conjugate base to the acid.
Here's how you can solve it step by step:
1. Convert the volume of the solution from milliliters to liters: 500 ml = 500/1000 = 0.5 L.
2. Calculate the concentration of the ethanoic acid in the solution. You have 0.1 mole of ethanoic acid dissolved in 0.5 L of water, so the concentration of the acid is:
Concentration (CH3COOH) = moles/volume
= 0.1 mol / 0.5 L
= 0.2 M (Molarity)
3. Calculate the concentration of the ethanoate (CH3COO-) ion, which is the conjugate base of ethanoic acid. Since the acid is monoprotic (donates one proton), the concentration of the conjugate base is equal to the concentration of the acid:
Concentration (CH3COO-) = 0.2 M
4. Calculate the ratio of [Base]/[Acid]:
[CH3COO-]/[CH3COOH] = 0.2 M / 0.2 M
= 1
5. Substitute the values into the Henderson-Hasselbalch equation:
pH = pKa + log([Base]/[Acid])
= 4.7 + log(1)
= 4.7 + 0
= 4.7
So, the pH of the solution is 4.7.
Remember, the Henderson-Hasselbalch equation assumes that the acid and its conjugate base are in equilibrium. Additionally, the equation is most accurate for weak acids and bases. In this example, ethanoic acid is a weak acid, so the Henderson-Hasselbalch equation is appropriate to use.