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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 described as strong because it completely ionizes in aqueous solution, meaning that all the acid molecules dissociate into ions. For example, hydrochloric acid (HCl) dissociates entirely into H⁺ and Cl⁻ ions, contributing to its strength compared to weak acids that only partially ionize.
It is also described as dilute because it has a low concentration, which indicates that there is a small amount of acid per unit volume of solution. This dilution does not affect the strength of the acid, as it remains fully ionized in the solution.
Step 2
What is the pH of a 1.0 × 10⁻⁵ mol/dm³ solution of hydrochloric acid?
Answer
To calculate the pH of a 1.0 × 10⁻⁵ mol/dm³ solution of hydrochloric acid, we start with the definition of pH:
Since hydrochloric acid is a strong acid, the concentration of hydrogen ions [H⁺] is equal to the concentration of the acid. Therefore,
Step 3
Calculate the concentration of the sulfuric acid in mol/dm³
Answer
From the titration data, we take the concordant results from the table: 22.13 cm³ and 22.15 cm³.
First, calculate the average titre:
ext{Average titre} = rac{22.13 + 22.15}{2} = 22.14 ext{ cm}³
Next, using the equation:
ightarrow ext{ Na}_2 ext{SO}_4 + 2 ext{ H}_2 ext{O}$$ The moles of sodium hydroxide used: $$ ext{moles NaOH} = rac{22.14}{1000} imes 0.105 = 0.00232 ext{ moles}$$ Based on the stoichiometry, 2 moles of NaOH react with 1 mole of H₂SO₄, thus: $$ ext{moles H}_2 ext{SO}_4 = rac{0.00232}{2} = 0.00116 ext{ moles}$$ The concentration of sulfuric acid: $$ ext{Concentration} = rac{0.00116 ext{ moles}}{0.025 ext{ dm}^3} = 0.0464 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 is designed to accurately measure a fixed volume of liquid, which ensures that the amount of dilute sulfuric acid used in each titration is consistent. This accuracy is critical to achieve reliable and reproducible results.
On the other hand, a burette provides variable volume measurements and allows for precise delivery of the sodium hydroxide solution dropwise. This is essential for determining the endpoint of the titration accurately.
Step 5
Calculate the mass of sodium hydroxide in 30.0 cm³ of a 0.105 mol/dm³ solution.
Answer
To calculate the mass of sodium hydroxide, we first find the number of moles:
ext{moles} = ext{concentration} imes ext{volume} = 0.105 ext{ mol/dm}^3 imes rac{30.0}{1000} ext{ dm}^3 = 0.00315 ext{ mol}
Next, using the relative formula mass of NaOH (40 g/mol):