Titration Calculations: Example Problems and Walkthrough

🧪 Quick Study Guide

• ⚗️ Titration is a process used to determine the concentration of a solution (analyte) by reacting it with a solution of known concentration (titrant).
• 🧮 The key formula for titration calculations is: $M_1V_1 = M_2V_2$, where:
  • $M_1$ = Molarity of the titrant
  • $V_1$ = Volume of the titrant
  • $M_2$ = Molarity of the analyte
  • $V_2$ = Volume of the analyte
• ⚖️ For reactions involving stoichiometry other than 1:1, adjust the formula to account for the mole ratios:
  • $\frac{M_1V_1}{n_1} = \frac{M_2V_2}{n_2}$
  • Where $n_1$ and $n_2$ are the stoichiometric coefficients from the balanced chemical equation.
• 🌡️ Equivalence Point: The point in the titration where the titrant has completely neutralized the analyte.
• indicator: A substance that changes color near the equivalence point, signaling the endpoint of the titration.
• 💡 Always balance the chemical equation before performing any calculations to ensure correct mole ratios.

📝 Practice Quiz

1. Question 1: 25.0 mL of a 0.10 M NaOH solution is required to neutralize 20.0 mL of an unknown HCl solution. What is the molarity of the HCl solution?
   1. 0.08 M
   2. 0.125 M
   3. 0.20 M
   4. 0.25 M
2. Question 2: In a titration, 30.0 mL of 0.20 M $H_2SO_4$ is needed to neutralize 40.0 mL of a KOH solution. What is the molarity of the KOH solution?
   1. 0.15 M
   2. 0.20 M
   3. 0.30 M
   4. 0.40 M
3. Question 3: What volume of 0.15 M HCl is required to neutralize 15.0 mL of 0.20 M $Ba(OH)_2$ solution?
   1. 10.0 mL
   2. 20.0 mL
   3. 30.0 mL
   4. 40.0 mL
4. Question 4: 10.0 mL of acetic acid ($CH_3COOH$) is titrated with 0.1 M NaOH. At the equivalence point, 15.0 mL of NaOH has been added. What is the molarity of the acetic acid solution?
   1. 0.067 M
   2. 0.10 M
   3. 0.15 M
   4. 0.20 M
5. Question 5: If 20.0 mL of 0.1 M $HBr$ is used to titrate 25.0 mL of $NH_3$, what is the concentration of the $NH_3$ solution?
   1. 0.06 M
   2. 0.08 M
   3. 0.10 M
   4. 0.125 M
6. Question 6: A 50.0 mL solution of $Ca(OH)_2$ is titrated with 0.2 M HCl. If 40.0 mL of HCl is required to reach the equivalence point, what is the molarity of the $Ca(OH)_2$ solution?
   1. 0.08 M
   2. 0.16 M
   3. 0.20 M
   4. 0.40 M
7. Question 7: What mass of KHP (potassium hydrogen phthalate, molar mass = 204.22 g/mol) is needed to neutralize 20.0 mL of 0.1 M NaOH?
   1. 0.204 g
   2. 0.408 g
   3. 0.612 g
   4. 0.816 g