A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture. After the reaction was complete, the resulting solution was transferred to a volumetric flask, made up to 250 cm³ with deionized water and mixed thoroughly. Several 25.0 cm³ portions of the resulting solution were titrated with 0.150 mol dm⁻³ aqueous sodium hydroxide. The mean titre was 26.60 cm³ of aqueous sodium hydroxide. Calculate the value of x in Na₂CO₃·xH₂O Show your working. Give your answer as an integer.

A-Level AQA Chemistry The Mole, Avogadro & The Ideal Gas Equation: A student added 627 mg of hydrat

A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture - AQA - A-Level Chemistry - Question 10 - 2018 - Paper 1

Question 10

A-student-added-627-mg-of-hydrated-sodium-carbonate-(Na₂CO₃·xH₂O)-to-200-cm³-of-0.250-mol-dm⁻³-hydrochloric-acid-in-a-beaker-and-stirred-the-mixture-AQA-A-Level Chemistry-Question 10-2018-Paper 1.png

A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture. After the react... show full transcript

Worked Solution & Example Answer:A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture - AQA - A-Level Chemistry - Question 10 - 2018 - Paper 1

Step 1

HCl added = 0.250 mol dm⁻³

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Answer

The volume of HCl used is 200 cm³, which is equivalent to 0.200 dm³. The moles of HCl can be calculated as:

$ext{Moles of HCl} = ext{Concentration} imes ext{Volume} = 0.250 ext{ mol dm}^{-3} imes 0.200 ext{ dm}^{3} = 0.050 ext{ mol}$

Step 2

NaOH used in titration = 3.99 x 10⁻² mol

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Answer

To find the moles of NaOH used to neutralize the remaining HCl after the reaction with Na₂CO₃·xH₂O, we can use the mean titre value of 26.60 cm³, converted to dm³:

$ext{Volume of NaOH} = 26.60 ext{ cm}^{3} = 0.0266 ext{ dm}^{3}$

The moles of NaOH used is then:

$ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.150 ext{ mol dm}^{-3} imes 0.0266 ext{ dm}^{3} = 0.00399 ext{ mol}$

Step 3

Therefore the moles of HCl reacted with the Na₂CO₃·xH₂O

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Answer

The moles of HCl that reacted with Na₂CO₃·xH₂O is:

$ext{Moles of HCl reacted} = ext{Initial moles of HCl} - ext{Moles of NaOH} = 0.050 - 0.00399 = 0.04601 ext{ mol}$

Step 4

5x moles Na₂CO₃·xH₂O reacted with the HCl

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Answer

From the reaction stoichiometry:

ightarrow 2 ext{NaCl} + ext{CO₂} + ext{H₂O}$$ 1 mole of Na₂CO₃\cdotxH₂O reacts with 2 moles of HCl. Therefore: $$5x = rac{0.04601}{2} = 0.023005 ext{ mol}$$

Step 5

Conversion of mg to g

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Answer

To calculate the moles of Na₂CO₃·xH₂O used:

$ext{Mass of Na₂CO₃·xH₂O} = 627 ext{ mg} = 0.627 ext{ g}$

Molar mass of Na₂CO₃ is approximately 106 g/mol. This gives:

$ext{Moles of Na₂CO₃\cdotxH₂O} = rac{0.627}{106} = 0.00591 ext{ mol}$

Step 6

Final calculation for x

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Answer

Combining the previous results:

$5x = 0.00591 ext{ mol}$

Thus,

$x = rac{0.00591}{5} = 0.001182 ext{ mol}$

Since each mole corresponds to 18 grams of water:

$ext{Value of } x = 1$

A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture - AQA - A-Level Chemistry - Question 10 - 2018 - Paper 1

Worked Solution & Example Answer:A student added 627 mg of hydrated sodium carbonate (Na₂CO₃·xH₂O) to 200 cm³ of 0.250 mol dm⁻³ hydrochloric acid in a beaker and stirred the mixture - AQA - A-Level Chemistry - Question 10 - 2018 - Paper 1

HCl added = 0.250 mol dm⁻³

NaOH used in titration = 3.99 x 10⁻² mol

Therefore the moles of HCl reacted with the Na₂CO₃·xH₂O

5x moles Na₂CO₃·xH₂O reacted with the HCl

Conversion of mg to g

Final calculation for x

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