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 created 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 ionises in water, thus classifying it as a strong acid.

Step 2

Which solution would have the lowest pH?

99%

104 rated

Answer

The solution with the lowest pH is the 1.0 mol/dm³ hydrogen chloride solution. Strong acids like hydrogen chloride fully ionise, leading to lower pH values compared to that of weak acids.

Step 3

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

96%

101 rated

Answer

  1. Use a more precise measuring instrument, such as a digital burette, to reduce human error in reading the volume of the ethanoic acid solution used.
  2. Perform the titration multiple times and calculate an average volume to minimize random errors in measurement.

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 of ethanoic acid needed:

  1. Calculate the number of moles required:

    moles=0.0480 mol/dm3×2501000 dm3=0.012 mol\text{moles} = 0.0480 \text{ mol/dm}^3 \times \frac{250}{1000} \text{ dm}^3 = 0.012 \text{ mol}

  2. Then, use the molar mass to find the mass:

    Mass=0.012 mol×90 g/mol=1.08 g\text{Mass} = 0.012 \text{ mol} \times 90 \text{ g/mol} = 1.08 \text{ g}

Step 5

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

97%

117 rated

Answer

Using the relation between moles and volume:

  1. Calculate the moles of ethanoic acid:

    moles H₂C₂O₄=0.0480 mol/dm3×15.001000 dm3=0.00072 mol\text{moles H₂C₂O₄} = 0.0480 \text{ mol/dm}^3 \times \frac{15.00}{1000} \text{ dm}^3 = 0.00072 \text{ mol}

  2. From the balanced equation, the moles of sodium hydroxide are:

    moles NaOH=0.00072 mol×2=0.00144 mol\text{moles NaOH} = 0.00072 \text{ mol} \times 2 = 0.00144 \text{ mol}

  3. Finally, calculate the concentration:

    Concentration=0.00144 mol0.0250 dm3=0.0576 mol/dm3\text{Concentration} = \frac{0.00144 \text{ mol}}{0.0250 \text{ dm}^3} = 0.0576 \text{ 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

;