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

"among the transition metal elements, manganese and zinc have a slight lowering of electronegativities. the reason behind this is that the nuclear charges of these elements are slightly weaker than the other transition metal elements due to the way their valence electrons are filled up. both manganese and zinc have their valence electrons filled up to the 4th shell with 1 pair of electrons (4s2) completely occupying it. although the electrons in the 3rd shell (3d) are yet to be paired since the outer most shell is filled up, the ability of the nucleus to attract electrons is lessened. hence,   the electronegativity (or the ability to attract electrons) is lessened."