Final answer:
After neutralizing the NaOH with HCl, 0.0015 moles of HCl remain in the mixture. The resulting concentration of HCl is 0.0429 M and the pH of the solution is 1.37.
Step-by-step explanation:
To find the pH of the solution after mixing 10 mL of 0.1 M NaOH with 25 mL of 0.1 M HCl, we must first determine if the acid and base completely neutralize each other.
Amount of NaOH = 0.01 L × 0.1 mol/L = 0.001 moles
Amount of HCl = 0.025 L × 0.1 mol/L = 0.0025 moles
Since NaOH and HCl react in a 1:1 molar ratio, and there is more HCl, the NaOH is the limiting reagent and will be completely neutralized by HCl:
0.0025 moles HCl - 0.001 moles NaOH = 0.0015 moles HCl remaining
The total volume of the solution after mixing is 10 mL + 25 mL = 35 mL or 0.035 L.
The concentration of HCl remaining is:
0.0015 moles / 0.035 L = 0.0429 M
The pH can then be calculated using the equation:
pH = -log[H+] = -log[0.0429] = 1.37
Therefore, the pH of the solution after mixing the two is 1.37, which corresponds to option a.