Titration Problem Set PDF | PDF | Acid | Titration - Free Printable

Let's solve each problem step-by-step using principles of stoichiometry and titration.

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Problem 1:

What is the M of NaOH if it takes 40.0 mL of NaOH to reach the equivalence point in a titration with 50.0 mL of 0.200 M HCl?

Step 1: Write the balanced equation:
\[
\text{HCl} + \text{NaOH} \rightarrow \text{NaCl} + \text{H}_2\text{O}
\]
1 mole HCl reacts with 1 mole NaOH.

Step 2: Use the formula:
\[
M_1 V_1 = M_2 V_2
\]
Where:
- \(M_1 = 0.200\) M (HCl)
- \(V_1 = 50.0\) mL
- \(V_2 = 40.0\) mL
- \(M_2 = ?\) (NaOH)

\[
(0.200)(50.0) = M_2 (40.0)
\]
\[
10.0 = 40.0 M_2 \Rightarrow M_2 = \frac{10.0}{40.0} = 0.250 \, \text{M}
\]

✔ Answer: 0.250 M NaOH

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Problem 2:

50. mL of 0.30 M KOH are required to titrate 60. mL of H₂SO₄. What is the M of H₂SO₄?

Balanced equation:
\[
\text{H}_2\text{SO}_4 + 2\text{KOH} \rightarrow \text{K}_2\text{SO}_4 + 2\text{H}_2\text{O}
\]
1 mole H₂SO₄ reacts with 2 moles KOH.

So we use:
\[
\text{Moles of KOH} = M \times V = 0.30 \, \text{M} \times 0.050 \, \text{L} = 0.015 \, \text{moles}
\]

From stoichiometry:
\[
1 \, \text{mol H}_2\text{SO}_4 : 2 \, \text{mol KOH}
\Rightarrow \text{Moles of H}_2\text{SO}_4 = \frac{0.015}{2} = 0.0075 \, \text{mol}
\]

Volume of H₂SO₄ = 60. mL = 0.060 L

\[
\text{Molarity} = \frac{0.0075}{0.060} = 0.125 \, \text{M}
\]

✔ Answer: 0.125 M H₂SO₄

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Problem 3:

60.0 mL of 1.20 M NaOH are required to titrate 40.0 mL of HF. What is the M of HF?

Balanced equation:
\[
\text{HF} + \text{NaOH} \rightarrow \text{NaF} + \text{H}_2\text{O}
\]
1:1 ratio.

Moles of NaOH = \(1.20 \, \text{M} \times 0.0600 \, \text{L} = 0.072 \, \text{mol}\)

So moles of HF = 0.072 mol

Volume of HF = 40.0 mL = 0.0400 L

\[
\text{M of HF} = \frac{0.072}{0.0400} = 1.80 \, \text{M}
\]

✔ Answer: 1.80 M HF

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Problem 4:

What volume of 0.40 M NaOH would be required to titrate 100. mL of 0.25 M HCl?

1:1 reaction: HCl + NaOH → NaCl + H₂O

Moles of HCl = \(0.25 \, \text{M} \times 0.100 \, \text{L} = 0.025 \, \text{mol}\)

So moles of NaOH needed = 0.025 mol

Volume of NaOH = \(\frac{0.025}{0.40} = 0.0625 \, \text{L} = 62.5 \, \text{mL}\)

✔ Answer: 62.5 mL

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Problem 5:

40.0 mL of 0.100 M H₃PO₄ are required to titrate 150.0 mL of NaOH to the equivalence point. What is the M of the NaOH?

Balanced equation:
\[
\text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}
\]
1 mole H₃PO₄ reacts with 3 moles NaOH.

Moles of H₃PO₄ = \(0.100 \, \text{M} \times 0.0400 \, \text{L} = 0.00400 \, \text{mol}\)

Moles of NaOH = \(3 \times 0.00400 = 0.0120 \, \text{mol}\)

Volume of NaOH = 150.0 mL = 0.150 L

\[
\text{M of NaOH} = \frac{0.0120}{0.150} = 0.0800 \, \text{M}
\]

✔ Answer: 0.0800 M NaOH

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Problem 6:

55 mL of 1.2 M HCl₂H₂O₂ are used to titrate a sample of 0.67 M Ba(OH)₂. What is the volume of the Ba(OH)₂ used?

Wait — HCl₂H₂O₂ is not a standard compound. This appears to be a typo.

Possibility: It may mean H₂C₂O₄ (oxalic acid), or perhaps H₂SO₄? But written as HCl₂H₂O₂?

Alternatively, could be H₂C₂O₄, oxalic acid, which is often written as H₂C₂O₄.

But let’s suppose it’s H₂C₂O₄ (oxalic acid), since HCl₂H₂O₂ doesn’t make sense.

Assuming H₂C₂O₄ (oxalic acid), a diprotic acid.

Reaction:
\[
\text{H}_2\text{C}_2\text{O}_4 + \text{Ba(OH)}_2 \rightarrow \text{BaC}_2\text{O}_4 + 2\text{H}_2\text{O}
\]
1:1 mole ratio.

But oxalic acid has two H⁺, Ba(OH)₂ has two OH⁻ → so yes, 1:1.

So:

Moles of H₂C₂O₄ = \(1.2 \, \text{M} \times 0.055 \, \text{L} = 0.066 \, \text{mol}\)

So moles of Ba(OH)₂ = 0.066 mol

M of Ba(OH)₂ = 0.67 M

Volume = \(\frac{0.066}{0.67} = 0.0985 \, \text{L} = 98.5 \, \text{mL}\)

✔ Answer: ~98.5 mL of Ba(OH)₂

> ⚠️ Note: If "HCl₂H₂O₂" was meant to be something else (like H₂SO₄), please clarify. But assuming it's a typo for oxalic acid.

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Problem 7:

90.0 mL of 0.255 M Ca(OH)₂ are required to titrate 100.0 mL of HCl. What is M of HCl?

Balanced equation:
\[
2\text{HCl} + \text{Ca(OH)}_2 \rightarrow \text{CaCl}_2 + 2\text{H}_2\text{O}
\]
2 HCl : 1 Ca(OH)₂

Moles of Ca(OH)₂ = \(0.255 \, \text{M} \times 0.0900 \, \text{L} = 0.02295 \, \text{mol}\)

So moles of HCl = \(2 \times 0.02295 = 0.0459 \, \text{mol}\)

Volume of HCl = 100.0 mL = 0.100 L

\[
\text{M of HCl} = \frac{0.0459}{0.100} = 0.459 \, \text{M}
\]

✔ Answer: 0.459 M HCl

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Problem 8:

50.2 mL of 0.453 M Sr(OH)₂ are required to titrate a 0.755 M H₂SO₄ sample. What is the volume of the H₂SO₄?

Balanced equation:
\[
\text{H}_2\text{SO}_4 + \text{Sr(OH)}_2 \rightarrow \text{SrSO}_4 + 2\text{H}_2\text{O}
\]
1:1 ratio.

Moles of Sr(OH)₂ = \(0.453 \, \text{M} \times 0.0502 \, \text{L} = 0.0227306 \, \text{mol}\)

So moles of H₂SO₄ = 0.0227306 mol

M of H₂SO₄ = 0.755 M

Volume = \(\frac{0.0227306}{0.755} = 0.03011 \, \text{L} = 30.11 \, \text{mL}\)

✔ Answer: 30.1 mL H₂SO₄

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Problem 9:

Would it take more 0.10 M HCl or 0.10 M H₂SO₄ to neutralize 30. mL of NaOH? Prove it!

We are comparing volumes of 0.10 M HCl vs 0.10 M H₂SO₄ to neutralize 30. mL of NaOH.

Let’s assume NaOH concentration is not given — but we can compare based on number of H⁺ ions per molecule.

Let’s suppose NaOH is 1.0 M (but actually, it's not specified). Wait — no, we need to know how much base we have.

But the question is asking which acid requires more volume to neutralize the same amount of NaOH.

So let’s say we have 30. mL of NaOH, but its molarity is not given — this is ambiguous.

But likely, it means same concentration of NaOH, say x M, and we’re comparing volumes of 0.10 M HCl and 0.10 M H₂SO₄ to neutralize it.

But without knowing the concentration of NaOH, we cannot compute exact volumes.

Wait — perhaps the idea is: Which acid, at the same molarity, will require more volume to neutralize the same amount of NaOH?

But that depends on proton count.

Let’s assume the NaOH solution is 1.0 M, for example.

Then moles of NaOH = \(1.0 \, \text{M} \times 0.030 \, \text{L} = 0.030 \, \text{mol}\)

Now:

- For HCl: monoprotic → 1 H⁺ per molecule.
- Moles of HCl needed = 0.030 mol
- Volume = \(\frac{0.030}{0.10} = 0.30 \, \text{L} = 300 \, \text{mL}\)

- For H₂SO₄: diprotic → 2 H⁺ per molecule.
- Moles of H₂SO₄ needed = \(0.030 / 2 = 0.015 \, \text{mol}\)
- Volume = \(\frac{0.015}{0.10} = 0.15 \, \text{L} = 150 \, \text{mL}\)

So HCl requires more volume than H₂SO₄.

✔ Answer: It would take more 0.10 M HCl than 0.10 M H₂SO₄ to neutralize 30 mL of NaOH.

Because H₂SO₄ provides two protons per molecule, so less volume is needed.

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Problem 10:

30.3 mL of 0.305 M NaOH are required to titrate H₃PO₄ to the equivalence point. How many moles of H₃PO₄ are needed to reach the equivalence point?

We need to know the stoichiometry.

H₃PO₄ is triprotic, but depending on the endpoint, it may not fully deprotonate.

But the question says "to the equivalence point", and does not specify which one.

However, unless stated otherwise, equivalence point typically refers to complete neutralization to PO₄³⁻.

So:

\[
\text{H}_3\text{PO}_4 + 3\text{NaOH} \rightarrow \text{Na}_3\text{PO}_4 + 3\text{H}_2\text{O}
\]

So 1 mole H₃PO₄ : 3 moles NaOH

Moles of NaOH = \(0.305 \, \text{M} \times 0.0303 \, \text{L} = 0.0092415 \, \text{mol}\)

So moles of H₃PO₄ = \(\frac{0.0092415}{3} = 0.0030805 \, \text{mol}\)

Rounded to 3 significant figures: 0.00308 mol

✔ Answer: 0.00308 moles of H₃PO₄

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## ✔ Final Answers Summary:

| Problem | Answer |
|--------|--------|
| 1 | 0.250 M NaOH |
| 2 | 0.125 M H₂SO₄ |
| 3 | 1.80 M HF |
| 4 | 62.5 mL NaOH |
| 5 | 0.0800 M NaOH |
| 6 | 98.5 mL Ba(OH)₂ *(assuming H₂C₂O₄)* |
| 7 | 0.459 M HCl |
| 8 | 30.1 mL H₂SO₄ |
| 9 | More 0.10 M HCl is needed; because H₂SO₄ is diprotic, fewer moles needed |
| 10 | 0.00308 mol H₃PO₄ |

Let me know if you'd like a visual explanation or diagram for any of these!