Final answer:
When 10 mL of 5.0 M HCl is added to 1 L of water, the pH can be calculated as 1.306. When 10 mL of 5 NAOH is added to 1 L of water, the pH can be calculated as 12.694.
Step-by-step explanation:
When 10 mL of 5.0 M HCl is added to 1 L of water, you are essentially diluting the HCl. To calculate the pH, you need to use the equation -log[H+]. In this case, [H+] can be calculated by dividing the moles of HCl by the total volume in liters. So, the pH would be -log(0.05 M/1.01 L) which is approximately -log(0.0495) = 1.306.
When 10 mL of 5 NAOH is added to 1 L of water, you are introducing OH- ions. The number of moles of OH- can be calculated by multiplying the molarity of the solution by the volume in liters. So, the moles of OH- would be (5 mol/L) * (0.01 L) = 0.05 mol. Since the OH- ions react with water to form H2O and increase the concentration of H+, the OH- acts as a base. To calculate the pH, you need to calculate the pOH using the equation -log[OH-]. In this case, the pOH would be -log(0.05 M/1.01 L) which is approximately -log(0.0495) = 1.306. The pH can be calculated by subtracting the pOH from 14, so the pH would be 14 - 1.306 = 12.694.