Final answer:
The pH of a solution prepared by adding 15 mL of 0.15 M NaCl to 15 mL of 0.15 M HCl will be dominated by the strong acid, HCl. The concentration of HCl is effectively 0.075 M due to dilution, resulting in a pH of approximately 1.12.
Step-by-step explanation:
When you add 15 mL of 0.15 M NaCl to a container containing 15 mL of 0.15 M HCl, the pH of the solution will be determined by the HCl component, as NaCl is a neutral salt and does not affect the pH.
Since HCl is a strong acid, it fully dissociates in water to release H+ ions.
Therefore, the pH of the solution can be calculated by considering the concentration of the HCl only.
As the volume doubles but the moles of HCl stay the same, the concentration of HCl is effectively halved.
The new concentration of HCl is 0.075 M.
The pH of the solution is calculated using the formula pH = -log[H+].
Therefore, the pH is approximately -log(0.075), which equals 1.12.