2025-2026 molar rations in titrataions

Last updated 20 days ago

10 Nsɛmmisa

2025-2026 molar rations in titrataions

Last updated 20 days ago

10 Nsɛmmisa

Work through the problems.

Then in your exercise book / notes

Write out the balanced equation for each titrataion and write the molar ratio below the acid and alkali.

Note if the ratio is 2:2 this will simplify to 1:1

Ba(OH)₂ + Na₂SO4 ⟶ BaSO₄ + 2H₂O

2 : 2

1 : 1

Hydrochloric Acid and Sodium Hydroxide

This is a classic strong acid-strong base titration.

Equation: null

Question: What is the molar ratio between hydrochloric acid (HCl) and sodium hydroxide (NaOH)?

Answer: The molar ratio is :. This means one mole of hydrochloric acid reacts completely with one mole of sodium hydroxide.

Mmuae Afoforo a Wobɛpaw:

Example 1: Calculating the Concentration of an Acid

A student finds that 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide (NaOH) solution is needed to neutralise 20.0 cm³ of a hydrochloric acid (HCI) solution. Calculate the concentration of the hydrochloric acid.

Step 1: Find the moles of the known substance (NaOH).
- Volume of NaOH = 25.0 cm³ = 0.025 dm³
- Moles of NaOH = Concentration × Volume
- Moles of NaOH = 0.100 mol/dm³ × 0.025 dm³ = 0.0025 mol
Step 2: Use the molar ratio to find the moles of the unknown substance (HCI).
- The ratio of HCI to NaOH is .
- Therefore, Moles of HCI = 0.0025 mol
Step 3: Calculate the concentration of the unknown substance (HCI).
- Concentration of HCI = Moles ÷ Volume
- Concentration of HCI = 0.0025 mol ÷ 0.020 dm³ = 0.125 mol/dm³

Mmuae Afoforo a Wobɛpaw:

1:1

2:1

1:2

Example 2: Calculating the Volume of an Alkali

How much 0.200 mol/dm³ sodium hydroxide (NaOH) solution is required to completely neutralise 15.0 cm³ of 0.150 mol/dm³ hydrochloric acid (HCI)?

Step 1: Find the moles of the known substance (HCI).
- Volume of HCI = 15.0 cm³ = 0.015 dm³
- Moles of HCI = 0.150 mol/dm³ × 0.015 dm³ = 0.00225 mol
Step 2: Use the molar ratio to find the moles of the unknown substance (NaOH).
- The ratio of HCI to NaOH is .
- Therefore, Moles of NaOH = 0.00225 mol
Step 3: Calculate the volume of the unknown substance (NaOH).
- Volume of NaOH = Moles ÷ Concentration
- Volume of NaOH = 0.00225 mol ÷ 0.200 mol/dm³ = 0.01125 dm³
- Volume in cm³ = 0.01125 × 1000 = 11.25 cm³

Mmuae Afoforo a Wobɛpaw:

2:1

1:2

1:1

Example 3: Calculating Concentration with Different Volumes

In a titration, 27.50 cm³ of 0.200 mol/dm³ hydrochloric acid (HCI) neutralised 25.00 cm³ of a sodium hydroxide (NaOH) solution. What is the concentration of the sodium hydroxide solution?

Step 1: Find the moles of HCI used.
- Volume of HCI = 27.50 cm³ = 0.0275 dm³
- Moles of HCI = 0.200 mol/dm³ × 0.0275 dm³ = 0.0055 mol
Step 2: Use the molar ratio to find the moles of NaOH.
- The ratio of HCI to NaOH is 1:1.
- Therefore, Moles of NaOH = 0.0055 mol
Step 3: Calculate the concentration of NaOH.
- Volume of NaOH = 25.00 cm³ = 0.025 dm³
- Concentration of NaOH = 0.0055 mol ÷ 0.025 dm³ = mol/dm³

Example 4: Calculating Concentration in g/dm³

A titration showed that 30.0 cm³ of sodium hydroxide (NaOH) was needed to neutralise 25.0 cm³ of 0.150 mol/dm³ hydrochloric acid (HCI). Calculate the concentration of the sodium hydroxide solution in grams per dm³ (g/dm³). (Relative atomic masses: Na = 23, O = 16, H = 1)

Step 1: Find the moles of HCI used.
- Volume of HCI = 25.0 cm³ = 0.025 dm³
- Moles of HCI = 0.150 mol/dm³ × 0.025 dm³ = 0.00375 mol
Step 2: Use the molar ratio to find the moles of NaOH.
- The ratio is 1:1, so Moles of NaOH = 0.00375 mol
Step 3: Calculate the concentration of NaOH in mol/dm³.
- Volume of NaOH = 30.0 cm³ = 0.030 dm³
- Concentration of NaOH = 0.00375 mol ÷ 0.030 dm³ = 0.125 mol/dm³
Step 4: Convert the concentration from mol/dm³ to g/dm³.
- First, find the molar mass (Mr) of NaOH: 23 + 16 + 1 = 40 g/mol
- Concentration (g/dm³) = Concentration (mol/dm³) × Mr
- Concentration (g/dm³) = 0.125 × 40 = g/dm³

A student needs to determine the concentration of a sulfuric acid (H₂SO₄) solution. They find that 20.0 cm³ of the acid is neutralised by 25.0 cm³ of 0.100 mol/dm³ sodium hydroxide (NaOH).

Equation: H₂SO₄ + 2NaOH ⟶ Na₂SO₄ + 2H₂O

Question: Calculate the concentration of the sulfuric acid solution.

Step 1: Calculate the moles of the known substance (NaOH).
- Volume = 25.0 cm³ = 0.025 dm³
- Moles of NaOH = Concentration × Volume
- Moles of NaOH = 0.100 mol/dm³ × 0.025 dm³ = 0.0025 mol
Step 2: Use the molar ratio to find the moles of the unknown substance (H₂SO₄).
- The ratio of H₂SO₄ to NaOH is .
- Moles of H₂SO₄ = Moles of NaOH ÷ 2
- Moles of H₂SO₄ = 0.0025 mol ÷ 2 = 0.00125 mol
Step 3: Calculate the concentration of the unknown substance (H₂SO₄).
- Volume = 20.0 cm³ = 0.020 dm³
- Concentration = Moles ÷ Volume
- Concentration = 0.00125 mol ÷ 0.020 dm³ = 0.0625 mol/dm³

25.0 cm³ of a calcium hydroxide (Ca(OH)₂) solution is titrated against 0.200 mol/dm³ hydrochloric acid (HCl). The endpoint is reached when 30.0 cm³ of HCl has been added.

Equation: 2HCl + Ca(OH)₂ ⟶ CaCl₂ + 2H₂O

Question: What is the concentration of the calcium hydroxide solution?

Step 1: Calculate the moles of HCl.
- Volume = 30.0 cm³ = 0.030 dm³
- Moles of HCl = 0.200 mol/dm³ × 0.030 dm³ = 0.006 mol
Step 2: Use the molar ratio to find the moles of Ca(OH)₂.
- The ratio of HCl to Ca(OH)₂ is .
- Moles of Ca(OH)₂ = Moles of HCl ÷ 2
- Moles of Ca(OH)₂ = 0.006 mol ÷ 2 = 0.003 mol
Step 3: Calculate the concentration of Ca(OH)₂.
- Volume = 25.0 cm³ = 0.025 dm³
- Concentration = 0.003 mol ÷ 0.025 dm³ = 0.120 mol/dm³

15.0 cm³ of a phosphoric acid (H₃PO₄) solution is completely neutralised by 35.0 cm³ of 0.150 mol/dm³ potassium hydroxide (KOH).

Equation: H₃PO₄ + 3KOH ⟶ K₃PO₄ + 3H₂O

Question: Calculate the concentration of the phosphoric acid.

Step 1: Calculate the moles of KOH.
- Volume = 35.0 cm³ = 0.035 dm³
- Moles of KOH = 0.150 mol/dm³ × 0.035 dm³ = 0.00525 mol
Step 2: Use the molar ratio to find the moles of H₃PO₄.
- The ratio of H₃PO₄ to KOH is .
- Moles of H₃PO₄ = Moles of KOH ÷ 3
- Moles of H₃PO₄ = 0.00525 mol ÷ 3 = 0.00175 mol
Step 3: Calculate the concentration of H₃PO₄.
- Volume = 15.0 cm³ = 0.015 dm³
- Concentration = 0.00175 mol ÷ 0.015 dm³ = 0.117 mol/dm³ (to 3 significant figures)

What volume of 0.500 mol/dm³ nitric acid (HNO₃) is required to neutralise 20.0 cm³ of 0.200 mol/dm³ barium hydroxide (Ba(OH)₂) solution?

Equation: 2HNO₃ + Ba(OH)₂ ⟶ Ba(NO₃)₂ + 2H₂O

Question: Determine the volume of nitric acid needed.

Step 1: Calculate the moles of Ba(OH)₂.
- Volume = 20.0 cm³ = 0.020 dm³
- Moles of Ba(OH)₂ = 0.200 mol/dm³ × 0.020 dm³ = 0.004 mol
Step 2: Use the molar ratio to find the moles of HNO₃.
- The ratio of HNO₃ to Ba(OH)₂ is .
- Moles of HNO₃ = Moles of Ba(OH)₂ × 2
- Moles of HNO₃ = 0.004 mol × 2 = 0.008 mol
Step 3: Calculate the volume of HNO₃ required.
- Volume = Moles ÷ Concentration
- Volume = 0.008 mol ÷ 0.500 mol/dm³ = 0.016 dm³
- Volume in cm³ = 0.016 × 1000 = 16.0 cm³

A chemist neutralises 50.0 cm³ of 0.250 mol/dm³ ammonia (NH₃) solution with sulfuric acid (H₂SO₄). The concentration of the sulfuric acid is 0.500 mol/dm³.

Equation: H₂SO₄ + 2NH₃ ⟶ (NH₄)₂SO₄

Question: Calculate the volume of sulfuric acid needed for the reaction.

Step 1: Calculate the moles of NH₃.
- Volume = 50.0 cm³ = 0.050 dm³
- Moles of NH₃ = 0.250 mol/dm³ × 0.050 dm³ = 0.0125 mol
Step 2: Use the molar ratio to find the moles of H₂SO₄.
- The ratio of H₂SO₄ to NH₃ is .
- Moles of H₂SO₄ = Moles of NH₃ ÷ 2
- Moles of H₂SO₄ = 0.0125 mol ÷ 2 = 0.00625 mol
Step 3: Calculate the volume of H₂SO₄ required.
- Volume = Moles ÷ Concentration
- Volume = 0.00625 mol ÷ 0.500 mol/dm³ = 0.0125 dm³
- Volume in cm³ = 0.0125 × 1000 = 12.5 cm³