Final answer:
After mixing 30.00 mL of 0.250 M HCl with 30.00 mL of 0.125 M NaOH, some HCl remains unreacted, making the solution acidic with a pH of approximately 1.2. The options provided in the question do not contain the correct pH value.
Step-by-step explanation:
The question relates to an acid-base neutralization reaction and the resultant pH after mixing solutions of HCl and NaOH. To find the pH of the solution obtained by mixing 30.00 mL of 0.250 M HCl with 30.00 mL of 0.125 M NaOH, we first calculate the moles of HCl and NaOH. For HCl, we have 30.00 mL × 0.250 M = 0.00750 moles, and for NaOH, we have 30.00 mL × 0.125 M = 0.00375 moles. The strong acid HCl will completely neutralize the same amount of the strong base NaOH. Hence, 0.00375 moles of HCl will be neutralized, leaving 0.00375 moles of HCl unreacted.
Since HCl is a strong acid, it will completely dissociate in water, so the concentration of remaining [H+] ions will be 0.00375 moles / 0.060 L = 0.0625 M. The pH is calculated as pH = -log([H+]) = -log(0.0625) = 1.2. Therefore, the final pH is neither neutral nor basic; it is acidic and close to option a. However, please note that option a, which states pH 0, is not the correct answer because the pH cannot be 0 in this scenario. None of the provided options b) 7, c) 9, or d) 13 are correct. The correct pH value is approximately 1.2.