This question is about acids - AQA - GCSE Chemistry - Question 9 - 2021 - Paper 1
Question 9
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 best 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 that would have the lowest pH is the 1.0 mol/dm³ hydrogen chloride solution. This is because it has the highest concentration of hydrogen ions due to the complete ionization of the strong acid.
Step 3
Suggest two improvements to the method that would increase the accuracy of the result.
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Use a digital burette: This allows for more precise measurement of the liquid being titrated, reducing human error in reading the scale.
Increase the number of trials: Performing more titrations and averaging the results can give a more accurate concentration value.
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, we use the formula: extMass=extConcentrationimesextVolumeimesextMr
Substituting in the values: ext{Mass} = 0.0480 ext{ mol/dm}^3 imes rac{250 ext{ cm}^3}{1000} imes 90
This results in a mass of: 4.32extg.
Step 5
Calculate the concentration of the sodium hydroxide solution in mol/dm³.
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Answer
From the balanced equation, 1 mole of H₂C₂O₄ reacts with 2 moles of NaOH.
The moles of H₂C₂O₄ used are calculated as follows: ext{moles H₂C₂O₄} = 0.0480 ext{ mol/dm}³ imes rac{15.00 ext{ cm}^3}{1000} = 0.00072 ext{ mol}
Thus, moles of NaOH = 0.00072 × 2 = 0.00144 mol.
Now, using the volume of NaOH (25.0 cm³ or 0.025 dm³): ext{Concentration of NaOH} = rac{0.00144 ext{ mol}}{0.025 ext{ dm}^3} = 0.0576 ext{ mol/dm}^3.