Question 10.12: Titration of an Acid A 25.0-mL (0.025 L) sample of an HCl so...
Titration of an Acid
A 25.0-mL (0.025 L) sample of an HCl solution is placed in a flask with a few drops of phenolphthalein (indicator). If 32.6 mL of a 0.185 M NaOH solution is needed to reach the endpoint, what is the molarity of the HCl solution?
NaOH(aq) + HCl(aq) \longrightarrow H_{2}O(l) + NaCl(aq)Step \fbox{1} State the given and needed quantities.
| Given | Need | |
| ANALYZE THE PROBLEM |
25.0 mL (0.0250 L) of HCl solution, | molarity of the HCl solution |
| 32.6 mL of 0.185 M NaOH solution | ||
| Neutralization Equation | ||
| NaOH(aq) + HCl(aq) \longrightarrow H_{2}O(l) + NaCl(aq) | ||
Step \fbox{2} Write a plan to calculate the molarity.
\begin{array}{r c} mL \text{ of } NaOH \\ \text{ solution }\end{array} \quad \begin{array}{r c} \text{ Metric }\\ \text{ factor }\end{array} \quad \begin{array}{r c} L \text{ of } NaOH \\ \text{ solution }\end{array} \quad \text{ Molarity } \quad \begin{array}{r c} \text{ moles of }\\ NaOH \end{array} \quad \begin{array}{r c} \text{ Mole–mole }\\ \text{ factor }\end{array} \quad \begin{array}{r c} \text{ moles of }\\ HCl \end{array} \quad \begin{array}{r c} \text{ Divide by }\\ \text{ liters } \end{array} \quad \begin{array}{r c} \text{ molarity of }\\ \text{ HCl solution }\end{array}Step \fbox{3} State equalities and conversion factors, including concentrations.
\begin{gathered}1 L \text { of } NaOH \text { solution }=1000 mL \text { of } NaOH \text { solution }\\\frac{1 L NaOH \text { solution }}{1000 mL NaOH \text { solution }} \quad \text { and } \quad \frac{1000 mL NaOH \text { solution }}{1 L NaOH \text { solution }} \quad \end{gathered} \begin{gathered}1 L \text { of } NaOH \text { solution }=0.185 \text { mole of } NaOH \\\frac{1 L NaOH \text { solution }}{0.185 \text { mole } NaOH} \quad \text { and } \quad \frac{0.185 \text { mole } NaOH}{1 L NaOH \text { solution }}\end{gathered}1 mole of HCl = 1 mole of NaOH
\frac{1 \text { mole } HCl }{ 1 \text { mole } NaOH} \quad \text { and } \quad \frac{ 1 \text { mole } NaOH }{1 \text { mole } HCl}Step \fbox{4} Set up the problem to calculate the needed quantity.
32.6 \cancel{mL NaOH \text { solution }} \times \frac{1 \cancel{L NaOH \text { solution }}}{1000 \cancel{mL NaOH \text { solution }}} \times \frac{0.185 \cancel{\text { mole } NaOH}}{1 \cancel{L NaOH \text { solution }}} \times \frac{1 \text { mole } HCl}{1 \cancel{\text { mole } NaOH}}=0.00603 \text{ mole of } HClmolarity of HCl = \frac{0.00603 \text{ mole } HCl}{0.0250 L \text { solution }} = 0.241 M HCl solution
Guide to Calculations for an Acid–Base Titration
Step \fbox{1}
State the given and needed quantities.
Step \fbox{2}
Write a plan to calculate the molarity or volume.
Step \fbox{3}
State equalities and conversion factors, including concentrations.
Step \fbox{4}
Set up the problem to calculate the needed quantity.