Calculate pH of a solution containing 0.1 mole of Ch3cooh, 0.2 mol of CH3COONa and 0.05 mol of naoh in 1 L. (Pka for Ch3cooh=4.74).
2. The attempt at a solution
The 0.05 mol of NaOH will react with the 0.10 mol CH3COOH to produce 0.05 mol CH3COONa , and there will bve 0.05 mol CH3COOH remaining unreacted . The solution then contains:
0.05 mol CH3COOH
0.25 mol CH3COONa.
dissolved in 1.0L solution - these figures are the molarity of the compounds.
Use Henderson - Hasselbalch equation:
pH = pKa + log ([salt]/[acid])
pH = 4.74 + log (0.25 / 0.05)
pH = 4.74 + log 58.0
pH = 4.74+ 0.70
pH = 5.44
I'm getting the correct answer this way but my question is that since some moles of ch3cooNa are already present in the beginning, won't that hinder more formation of the same salt and therefore the concentration of the salt calculated is actually more than the actual value?