86.14 mL of an acid solution was needed to neutralize 30.24 mL of a base solution of unknown concentrations. A second trial is run but this time 30.24 mL base solution is diluted to a total volume of 50.00 mL before starting the titration. How many mL of the acid solution are needed to neutralize it?
A) 45.26 mL
B) 86.14 mL
C) 61.11 mL
D) 52.10 mL

As given ,that 30.24 mL of base was neutralize by 86.14 mL of acid which means that moles of base present in 30.24 mL are neutralized by moles of acid present in 86.14 mL.

After dilution of base from 30.24 mL to 50.0 mL .Since, the moles of base are same in the solution as that of the moles in solution before dilution. Moles of acid require to neutralize the base after dilution will same as a that of present moles of acid present in 86.14 mL.

Oseni Oseni · Answer 2 · 2022-03-31T12:49:26+02:00

The total volume of the acid solution that would be needed to neutralize the base will remain 86.14 mL

Number of moles

From the illustration:

86.14 mL of the acid neutralizes 30.24 mL of the base. In other words, the number of moles of the acid and that of the neutralize one another.

Now, the same base is diluted to a volume of 50.00 mL. Both the volume and the concentration have changed but the number of moles remains the same.

Thus, the same number of moles of the same acid would still be required to neutralize the base. Thus, the volume of the acid required remains 86.14 mL

More on number of moles can be found here; https://brainly.com/question/14919968