Calculate the pH of the final solution when 25ml of ( N10) CH3COOH is titrated with(N10) NaOH solution:(i) 0 ml NaOH solution is added.(ii) 5 ml NaOH solution is added.(iii) 12.5 ml NaOH solution is added.(iv) 15 ml NaOH solution is added(v) 25 ml NaOH solution is added(vi) 30ml NaOH solution is added

To solve this problem, we need to understand that CH3COOH is a weak acid and NaOH is a strong base. When they react, they will form CH3COONa, a salt, and water. The pH of the solution will depend on the amount of NaOH added.

(i) 0 ml NaOH solution is added. Before any NaOH is added, the solution is just the weak acid CH3COOH. The pH of a weak acid is calculated using the formula pH = -log[H+], where [H+] is the concentration of hydrogen ions. For a 0.1 M solution of CH3COOH, the [H+] is approximately 0.00133 M (this can be found using the Ka of CH3COOH and the formula for the ionization of a weak acid). Therefore, the pH is approximately -log(0.00133) = 2.88.

(ii) 5 ml NaOH solution is added. When 5 ml of 0.1 M NaOH is added, it will react with 5 ml of the CH3COOH, leaving 20 ml of CH3COOH unreacted. The concentration of CH3COOH is now (20 ml / 30 ml) * 0.1 M = 0.067 M. Using the same method as above, the [H+] is approximately 0.00089 M and the pH is approximately -log(0.00089) = 3.05.

(iii) 12.5 ml NaOH solution is added. When 12.5 ml of 0.1 M NaOH is added, it will react with 12.5 ml of the CH3COOH, leaving 12.5 ml of CH3COOH unreacted. The concentration of CH3COOH is now (12.5 ml / 37.5 ml) * 0.1 M = 0.033 M. Using the same method as above, the [H+] is approximately 0.00061 M and the pH is approximately -log(0.00061) = 3.21.

(iv) 15 ml NaOH solution is added. When 15 ml of 0.1 M NaOH is added, it will react with 15 ml of the CH3COOH, leaving 10 ml of CH3COOH unreacted. The concentration of CH3COOH is now (10 ml / 40 ml) * 0.1 M = 0.025 M. Using the same method as above, the [H+] is approximately 0.00053 M and the pH is approximately -log(0.00053) = 3.28.

(v) 25 ml NaOH solution is added. When 25 ml of 0.1 M NaOH is added, it will react with all 25 ml of the CH3COOH, leaving no CH3COOH unreacted. The solution is now a solution of the salt CH3COONa, which is a basic salt. The pH will be greater than 7, but the exact value will depend on the Kb of CH3COONa.

(vi) 30 ml NaOH solution is added. When 30 ml of 0.1 M NaOH is added, it will react with all 25 ml of the CH3COOH and there will be 5 ml of NaOH unreacted. The solution is now a solution of NaOH, which is a strong base. The concentration of NaOH is (5 ml / 55 ml) * 0.1 M = 0.0091 M. The pOH is -log(0.0091) = 2.04, and the pH is 14 - pOH = 11.96.