35mL of acid with an unknown concentration is titrated to completion using 63mL of 3.0 MNaOH.

What is the concentration of the acid?

5.4 M.

Explanation:

At complete neutralization: It is known that the no. of millimoles of acid equal that of the base.

(MV) acid = (MV) NaOH

M of acid = ? M, V of acid = 35.0 mL.

M of NaOH = 3.0 M, V of NaOH = 63.0 mL.

∴ M of acid = (MV) NaOH / (V) acid = (3.0 M) (63.0 mL) / (35.0 mL) = 5.4 M.