From the above reaction, we can see that 1 mole of NaOH can neutralize the same amount of CH3COOH to produce 1 mole of CH3COONa.
Therefore, 6mL 0.15M NaOH can neutralize 6mL 0.15M CH3COOH to produce 6mL 0.15M CH3COONa
Now, remaining acid = (15 – 6)mL = 9mL
Produced salt = 6mL
Total volume of solution = (15 + 6)mL = 21mL