Titrations Practice Worksheet - Free Printable

Titrations Practice Worksheet Solutions

#### 1) If it takes 54 mL of 0.1 M NaOH to neutralize 125 mL of an HCl solution, what is the concentration of the HCl?

Solution:

The reaction between NaOH and HCl is a 1:1 molar ratio:
\[
\text{NaOH} + \text{HCl} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]

Using the titration equation:
\[
M_1 V_1 = M_2 V_2
\]
where:
- \( M_1 \) is the molarity of NaOH,
- \( V_1 \) is the volume of NaOH,
- \( M_2 \) is the molarity of HCl,
- \( V_2 \) is the volume of HCl.

Given:
- \( M_1 = 0.1 \, \text{M} \)
- \( V_1 = 54 \, \text{mL} \)
- \( V_2 = 125 \, \text{mL} \)

Rearrange the equation to solve for \( M_2 \):
\[
M_2 = \frac{M_1 V_1}{V_2}
\]

Substitute the values:
\[
M_2 = \frac{(0.1 \, \text{M})(54 \, \text{mL})}{125 \, \text{mL}}
\]

Calculate:
\[
M_2 = \frac{5.4}{125} = 0.0432 \, \text{M}
\]

Answer:
\[
\boxed{0.0432 \, \text{M}}
\]

---

#### 2) If it takes 25 mL of 0.05 M HCl to neutralize 345 mL of NaOH solution, what is the concentration of the NaOH solution?

Solution:

The reaction between HCl and NaOH is also a 1:1 molar ratio:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]

Using the titration equation:
\[
M_1 V_1 = M_2 V_2
\]
where:
- \( M_1 \) is the molarity of HCl,
- \( V_1 \) is the volume of HCl,
- \( M_2 \) is the molarity of NaOH,
- \( V_2 \) is the volume of NaOH.

Given:
- \( M_1 = 0.05 \, \text{M} \)
- \( V_1 = 25 \, \text{mL} \)
- \( V_2 = 345 \, \text{mL} \)

Rearrange the equation to solve for \( M_2 \):
\[
M_2 = \frac{M_1 V_1}{V_2}
\]

Substitute the values:
\[
M_2 = \frac{(0.05 \, \text{M})(25 \, \text{mL})}{345 \, \text{mL}}
\]

Calculate:
\[
M_2 = \frac{1.25}{345} \approx 0.00362 \, \text{M}
\]

Answer:
\[
\boxed{0.00362 \, \text{M}}
\]

---

#### 3) If it takes 50 mL of 0.5 M KOH solution to completely neutralize 125 mL of sulfuric acid solution (H₂SO₄), what is the concentration of the H₂SO₄ solution?

Solution:

The reaction between KOH and H₂SO₄ is:
\[
2 \text{KOH} + \text{H}_2\text{SO}_4 \rightarrow \text{K}_2\text{SO}_4 + 2 \text{H}_2\text{O}
\]

Here, the molar ratio of KOH to H₂SO₄ is 2:1. Therefore, we need to adjust the titration equation to account for this ratio:
\[
M_1 V_1 \times \text{stoichiometric factor} = M_2 V_2
\]

Given:
- \( M_1 = 0.5 \, \text{M} \) (KOH)
- \( V_1 = 50 \, \text{mL} \) (KOH)
- \( V_2 = 125 \, \text{mL} \) (H₂SO₄)
- Stoichiometric factor for H₂SO₄ = \( \frac{1}{2} \)

Rearrange the equation to solve for \( M_2 \):
\[
M_2 = \frac{M_1 V_1 \times \text{stoichiometric factor}}{V_2}
\]

Substitute the values:
\[
M_2 = \frac{(0.5 \, \text{M})(50 \, \text{mL}) \times \frac{1}{2}}{125 \, \text{mL}}
\]

Calculate:
\[
M_2 = \frac{(0.5)(50) \times 0.5}{125} = \frac{12.5}{125} = 0.1 \, \text{M}
\]

Answer:
\[
\boxed{0.1 \, \text{M}}
\]

---

#### 4) Can I titrate a solution of unknown concentration with another solution of unknown concentration and still get a meaningful answer? Explain your answer in a few sentences.

Solution:

No, you cannot titrate a solution of unknown concentration with another solution of unknown concentration and obtain a meaningful answer.

In a titration, the concentration of one solution (the titrant) must be known so that the concentration of the other solution (the analyte) can be calculated using the titration equation \( M_1 V_1 = M_2 V_2 \). If both concentrations are unknown, there is no reference point to determine the relationship between the volumes and concentrations, making it impossible to calculate either concentration accurately.

Answer:
\[
\boxed{\text{No, because at least one concentration must be known to calculate the other.}}
\]

---

#### 5) Explain the difference between an endpoint and equivalence point in a titration.

Solution:

- Equivalence Point: This is the point in a titration where the number of moles of the titrant added is chemically equivalent to the number of moles of the analyte present. At the equivalence point, the reaction is complete, and the stoichiometric amount of reactants has been reached. The equivalence point is determined by the chemistry of the reaction and is independent of any indicators or measurements.

- Endpoint: This is the point in a titration where the indicator changes color, signaling that the titration is complete. The endpoint is an experimental observation used to approximate the equivalence point. Ideally, the endpoint should coincide with the equivalence point, but due to the limitations of indicators, there may be a small discrepancy between the two.

In summary:
- The equivalence point is the true chemical point where the reaction is complete.
- The endpoint is the observed point where the indicator changes color, which is used to approximate the equivalence point.

Answer:
\[
\boxed{\text{The equivalence point is the exact point where the reaction is complete, while the endpoint is the observed point where the indicator changes color.}}
\]

---

Final Answers:

1. \(\boxed{0.0432 \, \text{M}}\)
2. \(\boxed{0.00362 \, \text{M}}\)
3. \(\boxed{0.1 \, \text{M}}\)
4. \(\boxed{\text{No, because at least one concentration must be known to calculate the other.}}\)
5. \(\boxed{\text{The equivalence point is the exact point where the reaction is complete, while the endpoint is the observed point where the indicator changes color.}}\)