35.0 ml of acid with an unknown concentration is titrated with 24.6 ml of 0.432 m base. what is the concentration of the acid? assume the acid contributes 1 mole of (h+) ions/mole of acid and the base contributes 1 mole of (oh-) ions/mole of base.

3.29 m
0.615 m
0.304 m

Given:

35.0 mL of acid with an unknown concentration

24.6 mL of 0.432 M base

Required:

Concentration of the acid

Solution:

M1V1 = M2V2

M1 (35.0 mL of acid) = (0.432 M base) (24.6 mL of base)

V1 = (0.432 M base) (24.6 mL of base) / (35.0 mL of acid)

M1 = 0.304 M of acid

i believe the answer is true.