Titrations Following are data for the titration of 0.108 M H2SO4 with a solution of NaOH of unknown concentration. What is the concentration of the NaOH solution? Strategy Use the volume of the acid and its concentration to calculate how many moles of hydrogen ions are available to be titrated. At

Question

Titrations

Following are data for the titration of 0.108 [latex]M H _{2} SO _{4}[/latex] with a solution of [latex]NaOH[/latex] of unknown concentration. What is the concentration of the [latex]NaOH[/latex] solution?

	Volume of 0.108 M [latex]H _{2} SO _{4}[/latex]	Volume of NaOH
Trial I	25.0 mL	33.48 mL
Trial II	25.0 mL	33.46 mL
Trial III	25.0 mL	33.50 mL

Strategy

Use the volume of the acid and its concentration to calculate how many moles of hydrogen ions are available to be titrated. At the equivalence point, the moles of base used will equal the moles of [latex]H ^{+}[/latex] available. Divide the moles of [latex]H ^{+}[/latex] by the volume of base used in liters to calculate the concentration of the base.

Accepted Answer

From the balanced equation for this acid–base reaction, we know the stoichiometry: Two moles of NaOH react with one mole of $H _{2} SO _{4}$ . From the three trials, we calculate that the average volume of the NaOH required for complete reaction is 33.48 mL. Because the units of molarity are moles/liter, we must convert volumes of reactants from milliliters to liters. We can then use the factor-label method (Section 1.5) to calculate the molarity of the NaOH solution. What we wish to calculate is the number of moles of NaOH per liter of NaOH.

\begin{aligned}\frac{ mol NaOH }{ L NaOH } &=\frac{0.108 mol H _{2} SO _{4}}{1 LH _{2} SO _{4}^{-}} \times \frac{0.0250 LH _{2} SO _{4}^{-}}{0.03348 L NaOH } \times \frac{2 mol NaOH }{1 mol _{2} SO } \\&=\frac{0.161 mol NaOH }{ L NaOH }=0.161 M\end{aligned}

[latex]\begin{aligned}\frac{ mol NaOH }{ L NaOH } &=\frac{0.108 mol H _{2} SO _{4}}{1 LH _{2} SO _{4}^{-}} imes \frac{0.0250 LH _{2} SO _{4}^{-}}{0.03348 L NaOH } imes \frac{2 mol NaOH }{1 mol _{2} SO } \&=\frac{0.161 mol NaOH }{ L NaOH }=0.161 M\end{aligned}[/latex]

Question 8.8: Titrations Following are data for the titration of 0.108 M H...

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