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 deionised 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 Formulae, Equations & Calculations: 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 reac... 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 concentration of hydrochloric acid used in the reaction is 0.250 mol dm⁻³, and the volume is 200 cm³. Therefore, we can calculate the amount of moles of HCl:

$ext{Moles of HCl} = ext{Concentration} imes ext{Volume} = 0.250 imes 0.200 = 0.050 ext{ mol}$

Step 2

NaOH used in titration = 0.150 mol dm⁻³

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Answer

We now need to calculate the total moles of NaOH used in the titration:

$ext{Mean titre} = 26.60 ext{ cm}^3 = 0.02660 ext{ dm}^3$

$ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.150 imes 0.02660 = 0.00399 ext{ mol}$

Step 3

Therefore the moles of HCl reacted with the Na2CO3

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Answer

From the reaction, the stoichiometric ratio of HCl to Na2CO3 is 2:1. Therefore, the moles of HCl that reacted with Na2CO3 is:

$ext{Moles of HCl reacted} = 0.00399 ext{ mol} imes 2 = 0.00798 ext{ mol}$

Step 4

Conversion of mg to moles

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The mass of hydrated sodium carbonate is given as 627 mg, which we can convert to grams:

$627 ext{ mg} = 0.627 ext{ g}$

To calculate the moles of Na2CO3, we need its molar mass. The formula for Na2CO3.xH2O can be written as:

$ext{Molar mass} = (2 imes 23) + 12 + (3 imes 16) + (x imes 18) = 46 + 12 + 48 + 18x = 60 + 18x$

Step 5

Moles of Na2CO3 reacted

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Answer

Using the calculated molar mass, we can establish the relationship:

$ext{Moles of Na2CO3} = rac{0.627}{60 + 18x}$

Setting this equal to the moles of HCl reacted:

$rac{0.627}{60 + 18x} = 0.00798$

Step 6

Solving for x

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Answer

Cross-multiplying to solve for x:

$0.627 = 0.00798(60 + 18x)$

This results in:

$0.627 = 0.4788 + 0.14364x$

Isolating x gives:

$0.627 - 0.4788 = 0.14364x$

$0.1482 = 0.14364x$

Thus,

ightarrow x ext{ is approximately } 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 = 0.150 mol dm⁻³

Therefore the moles of HCl reacted with the Na2CO3

Conversion of mg to moles

Moles of Na2CO3 reacted

Solving for x

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