Answer :
After adding 39.3 mL of 0.100 M NaOH to 50.0 mL of 0.200 M HClO₄, the pH is approximately 0.914, calculated using the remaining concentration of H⁺ ions.
To calculate the pH after adding 39.3 mL of 0.100 M NaOH to 50.0 mL of 0.200 M HClO₄, we first need to determine the number of moles of acid (HClO₄) initially present and the number of moles of base (NaOH) added.
Using the formula [tex]\(Molarity \times Volume = moles\)[/tex], we find the initial moles of HClO₄:
[tex]\(Moles_{HClO_4} = 0.200 \, M \times 0.050 \, L = 0.010 \, moles\)[/tex]
Next, we calculate the moles of NaOH added:
[tex]\(Moles_{NaOH} = 0.100 \, M \times 0.0393 \, L = 0.00393 \, moles\)[/tex]
Since NaOH is a strong base, it reacts completely with HClO₄ in a 1:1 ratio, forming water and NaClO₄. Therefore, the remaining moles of HClO₄ after the reaction equals the initial moles minus the moles of NaOH added:
[tex]\(Moles_{HClO_4, remaining} = 0.010 \, moles - 0.00393 \, moles = 0.00607 \, moles\)[/tex]
Now, we calculate the concentration of HClO₄ after the reaction:
[tex]\(Concentration_{HClO_4, remaining} = \frac{0.00607 \, moles}{0.050 \, L} = 0.1214 \, M\)[/tex]
Since HClO₄ is a strong acid, it completely dissociates in water to form H⁺ ions. Therefore, the concentration of H⁺ ions in the solution equals the concentration of remaining HClO₄:
[tex]\[ [H^+] = 0.1214 \, M \][/tex]
Finally, we calculate the pH using the formula [tex]\( pH = -\log[H^+] \):[/tex]
[tex]\[ pH = -\log(0.1214) \approx 0.914 \][/tex]
Therefore, the pH after adding 39.3 mL of NaOH is approximately 0.914.