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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

The solution made by dissolving 1 g of hydrogen chloride in 1 dm³ of water is a dilute solution of a strong acid. This is because hydrogen chloride completely ionizes in water and the concentration is low given the amount of solute compared to the solvent.

Step 2

Which solution would have the lowest pH?

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Answer

The solution with the lowest pH would be the 1.0 mol/dm³ hydrogen chloride solution. This is because it has the highest concentration of hydrogen ions, which results in a lower pH value.

Step 3

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

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Answer

  1. Use a white tile underneath the conical flask to easily see the color change of the indicator, ensuring more precise endpoint detection.
  2. Repeat the titration several times and calculate the average to improve reliability and account for any inconsistencies.

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 calculate the mass, first determine the number of moles required:

ext{Moles} = ext{Concentration} imes ext{Volume} = 0.0480 ext{ mol/dm}^3 imes rac{250 ext{ cm}^3}{1000} = 0.012 ext{ mol}

Now, use the relative formula mass to find the mass:

extMass=extMolesimesextMr=0.012extmolimes90extg/mol=1.08extg ext{Mass} = ext{Moles} imes ext{Mr} = 0.012 ext{ mol} imes 90 ext{ g/mol} = 1.08 ext{ g}

Step 5

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

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Answer

From the reaction, we know the ratio of moles of H₂C₂O₄ to NaOH is 1:2. Thus, the moles of sodium hydroxide are:

extMolesofNaOH=0.00072extmolimes2=0.00144extmol ext{Moles of NaOH} = 0.00072 ext{ mol} imes 2 = 0.00144 ext{ mol}

Now calculate the concentration:

ext{Concentration} = rac{ ext{Moles}}{ ext{Volume}} = rac{0.00144 ext{ mol}}{ rac{25.0}{1000} ext{ dm}^3} = 0.0576 ext{ mol/dm}^3

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