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This question is about acids and alkalis - AQA - GCSE Chemistry - Question 8 - 2022 - Paper 1

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This question is about acids and alkalis. 1. Explain why the pH of an acid depends on: • the strength of the acid: • the concentration of the acid. 2. A stud... show full transcript

Worked Solution & Example Answer:This question is about acids and alkalis - AQA - GCSE Chemistry - Question 8 - 2022 - Paper 1

Step 1

Explain why the pH of an acid depends on the strength of the acid:

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Answer

The pH of an acid is directly related to the concentration of hydrogen ions, H⁺, in the solution. The stronger the acid, the more it dissociates in aqueous solution to release H⁺ ions. Hence, stronger acids produce a higher concentration of H⁺ ions, leading to a lower pH value.

Step 2

Explain why the pH of an acid depends on the concentration of the acid:

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Answer

The pH is also affected by the concentration of the acid. A higher concentration of acid means a greater number of H⁺ ions present in the solution. Consequently, as the concentration increases, the pH decreases. Thus, a more concentrated acid will have a lower pH.

Step 3

Explain why the student used a volume of 23.50 cm³ of barium hydroxide solution in the titration calculation:

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Answer

The student calculated the mean volume from titration numbers 2 to 5, which were more accurate than the first titration. This was done to reduce anomalies, ensuring a reliable mean value was used for the titration calculation.

Step 4

Calculate the concentration of the hydrochloric acid in mol/dm³:

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Answer

To find the concentration of hydrochloric acid, we first calculate the moles of barium hydroxide used:

moles Ba(OH)2=23.501000×0.100=0.00235 mol\text{moles } Ba(OH)_2 = \frac{23.50}{1000} \times 0.100 = 0.00235 \text{ mol}

Since the mole ratio from the equation is 2:1 (2 moles of HCl for each mole of Ba(OH)₂), it follows that:

moles HCl=2×0.00235=0.00470 mol\text{moles } HCl = 2 \times 0.00235 = 0.00470 \text{ mol}

Next, the concentration is calculated as:

concentration =0.00470 mol25.00 cm3×1000=0.188 mol/dm3\text{concentration } = \frac{0.00470 \text{ mol}}{25.00 \text{ cm}^3} \times 1000 = 0.188 \text{ mol/dm}³

Step 5

Explain why the electrical conductivity of the mixture was zero when the sulfuric acid had just been neutralised:

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Answer

After neutralisation, the mixture contains solid barium sulfate, which is not soluble and does not dissociate into ions. Therefore, there are no free ions to conduct electricity in the solution, resulting in a conductivity of zero.

Step 6

Give one reason why the electrical conductivity of the mixture increased after adding a further 10 cm³ of barium hydroxide solution:

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Answer

Adding more barium hydroxide solution introduces additional hydroxide ions, OH⁻, into the solution. The presence of these free ions allows the solution to conduct electricity better, increasing the electrical conductivity.

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