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This question is about acids - AQA - GCSE Chemistry - Question 9 - 2021 - Paper 1

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This question is about acids. Hydrogen chloride and ethanoic acid both dissolve in water. All hydrogen chloride molecules ionise in water. Approximately 1% of ethan... show full transcript

Worked Solution & Example Answer:This question is about acids - AQA - GCSE Chemistry - Question 9 - 2021 - Paper 1

Step 1

Which is the correct description of this solution?

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Answer

This solution, made by dissolving 1 g of hydrogen chloride in 1 dm³ of water, is described as a dilute solution of a strong acid. This is due to the complete ionization of hydrogen chloride in water.

Step 2

Which solution would have the lowest pH?

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Answer

The solution with the lowest pH among the given options is the 1.0 mol/dm³ hydrogen chloride solution. Strong acids, such as hydrogen chloride, completely ionize, leading to a lower pH compared to the other solutions listed.

Step 3

Suggest two improvements to the method that would increase the accuracy of the result.

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Answer

  1. Swirling the solution during titration ensures thorough mixing of the reactants, leading to a more accurate endpoint.

  2. Using a white tile beneath the flask will improve visibility and make it easier to detect the endpoint changes when the indicator shifts color.

Step 4

Calculate the mass of ethanoic acid (H₂C₂O₄) needed to make 250 cm³ of a solution with concentration 0.0480 mol/dm³.

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Answer

To find the mass of ethanoic acid needed, we can use the formula:

extMass=extConcentrationimesextVolumeimesextMolarmass ext{Mass} = ext{Concentration} imes ext{Volume} imes ext{Molar mass}

Substituting the known values:

ext{Mass} = 0.0480 ext{ mol/dm}^3 imes rac{250 ext{ cm}^3}{1000} imes 90 ext{ g/mol} = 4.32 ext{ g}

Step 5

Calculate the concentration of the sodium hydroxide solution in mol/dm³.

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Answer

Using the balanced equation, the moles of ethanoic acid used can be calculated as:

ext{Moles of H₂C₂O₄} = 0.0480 rac{ ext{mol}}{ ext{dm}^3} imes rac{15.00 ext{ cm}^3}{1000} = 0.00072 ext{ mol}

From the stoichiometry of the reaction, the moles of NaOH would be:

extMolesofNaOH=2imes0.00072=0.00144extmol ext{Moles of NaOH} = 2 imes 0.00072 = 0.00144 ext{ mol}

Calculating the concentration:

ext{Concentration} = rac{0.00144 ext{ mol}}{0.0250 ext{ dm}^3} = 0.0576 ext{ mol/dm}^3

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