Photo AI

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

Question icon

Question 9

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

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?

96%

114 rated

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?

99%

104 rated

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.

96%

101 rated

Answer

  1. Use a digital burette: This allows for more precise measurement of the liquid being titrated, reducing human error in reading the scale.
  2. 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.

98%

120 rated

Answer

To find the mass, we use the formula:
extMass=extConcentrationimesextVolumeimesextMrext{Mass} = ext{Concentration} imes ext{Volume} imes ext{Mr}
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:
4.32extg4.32 ext{ g}.

Step 5

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

97%

117 rated

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.

Join the GCSE students using SimpleStudy...

97% of Students

Report Improved Results

98% of Students

Recommend to friends

100,000+

Students Supported

1 Million+

Questions answered

;