Question 10.14: In a sulfuric acid (H2SO4)–sodium hydroxide (NaOH) acid–base......

First, we calculate the number of moles of H_{2}SO_{4} that reacted with the NaOH. The pathway for this calculation, using dimensional analysis (Section 6.8), is
mL of NaOH → L of NaOH → moles of NaOH → moles of H_{2}SO_{4}
The sequence of conversion factors that effects this series of unit changes is

The first conversion factor derives from the definition of a milliliter, the second conversion factor derives from the definition of molarity (Section 8.5), and the third conversion factor uses the coefficients in the balanced chemical equation for the titration reaction (Section 6.7).
The number of moles of H_{2}SO_{4} that react is obtained by combining all the numbers in the dimensional analysis setup in the manner indicated.

(\frac{17.3\times 10^{-3}\times 0.126\times 1}{1\times 1\times 2} ) mole H_{2}SO_{4} = 0.00109 mole H_{2}SO_{4}

Now that we know how many moles of H_{2}SO_{4} reacted, we calculate the molarity of the H_{2}SO_{4} solution using the definition for molarity.

Molarity H_{2}SO_{4} = \frac{moles \ H_{2}SO_{4}}{L \ H_{2}SO_{4} \ solution} = \frac{0.00109 \ mole}{0.0250 \ L} \\ = 0.0436 \ \frac{mole}{L}

Note that the units in the denominator of the molarity equation must be liters (0.0250) rather than milliliters (25.0).