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This question is about acids and alkalis - AQA - GCSE Chemistry - Question 9 - 2018 - Paper 1
Question 9
This question is about acids and alkalis. Dilute hydrochloric acid is a strong acid. Explain why an acid can be described as both strong and dilute. A 1.0 × 10⁻³ ... show full transcript
Worked Solution & Example Answer:This question is about acids and alkalis - AQA - GCSE Chemistry - Question 9 - 2018 - Paper 1
Step 1
Explain why an acid can be described as both strong and dilute.
Answer
An acid is considered strong because it completely ionizes in aqueous solution. This means that all the acid molecules dissociate into ions, contributing to a higher concentration of hydrogen ions in the solution. On the other hand, an acid is described as dilute when it is present in a small amount per unit volume. This indicates that even though it is strong in terms of ionization, the concentration of the acid in the solution is low.
Step 2
Step 3
Calculate the concentration of the sulfuric acid in mol/dm³ using only the student’s concordant results.
Answer
The average titre from the concordant results (titration 3 and 4) is:
ext{Average titre} = rac{22.13 + 22.15}{2} = 22.14 ext{ cm}^3
Next, we find the moles of NaOH used:
ext{Moles of NaOH} = rac{22.14 ext{ cm}^3}{1000} imes 0.105 = 0.002325 ext{ mol}
Using the reaction stoichiometry (2 NaOH : 1 H₂SO₄), the moles of H₂SO₄ can be calculated as follows:
ext{Moles of H₂SO₄} = rac{0.002325}{2} = 0.0011625 ext{ mol}
Finally, to find the concentration of H₂SO₄:
ext{Concentration} = rac{0.0011625 ext{ mol}}{0.025 ext{ dm}^3} = 0.0465 ext{ mol/dm}^3
Step 4
Explain why the student should use a pipette to measure the dilute sulfuric acid and a burette to measure the sodium hydroxide solution.
Answer
A pipette measures a fixed volume accurately, ensuring that the exact amount of dilute sulfuric acid is delivered for the reaction. In contrast, a burette allows for variable volume measurements, enabling the student to add sodium hydroxide drop by drop until reaching the endpoint of the titration, which provides better control and precision in determining the exact volume used.
Step 5
Calculate the mass of sodium hydroxide in 30 cm³ of a 0.105 mol/dm³ solution.
Answer
First, we calculate the number of moles of sodium hydroxide:
ext{Moles} = ext{Concentration} imes ext{Volume} = 0.105 ext{ mol/dm}^3 imes rac{30 ext{ cm}^3}{1000} = 0.00315 ext{ mol}
Next, we calculate the mass using the relative formula mass (Mᵣ):