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Parents Pricing Home A-Level AQA Chemistry Formulae, Equations & Calculations 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
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
View full question 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
View marking scheme 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 Only available for registered users.
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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:
e x t M o l e s o f H C l = e x t C o n c e n t r a t i o n i m e s e x t V o l u m e = 0.250 i m e s 0.200 = 0.050 e x t m o l ext{Moles of HCl} = ext{Concentration} imes ext{Volume} = 0.250 imes 0.200 = 0.050 ext{ mol} e x t M o l eso f H Cl = e x t C o n ce n t r a t i o n im ese x t V o l u m e = 0.250 im es 0.200 = 0.050 e x t m o l
NaOH used in titration = 0.150 mol dm⁻³ Only available for registered users.
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We now need to calculate the total moles of NaOH used in the titration:
e x t M e a n t i t r e = 26.60 e x t c m 3 = 0.02660 e x t d m 3 ext{Mean titre} = 26.60 ext{ cm}^3 = 0.02660 ext{ dm}^3 e x t M e an t i t re = 26.60 e x t c m 3 = 0.02660 e x t d m 3
e x t M o l e s o f N a O H = e x t C o n c e n t r a t i o n i m e s e x t V o l u m e = 0.150 i m e s 0.02660 = 0.00399 e x t m o l ext{Moles of NaOH} = ext{Concentration} imes ext{Volume} = 0.150 imes 0.02660 = 0.00399 ext{ mol} e x t M o l eso f N a O H = e x t C o n ce n t r a t i o n im ese x t V o l u m e = 0.150 im es 0.02660 = 0.00399 e x t m o l
Therefore the moles of HCl reacted with the Na2CO3 Only available for registered users.
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From the reaction, the stoichiometric ratio of HCl to Na2CO3 is 2:1. Therefore, the moles of HCl that reacted with Na2CO3 is:
e x t M o l e s o f H C l r e a c t e d = 0.00399 e x t m o l i m e s 2 = 0.00798 e x t m o l ext{Moles of HCl reacted} = 0.00399 ext{ mol} imes 2 = 0.00798 ext{ mol} e x t M o l eso f H Cl re a c t e d = 0.00399 e x t m o l im es 2 = 0.00798 e x t m o l
Conversion of mg to moles Only available for registered users.
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The mass of hydrated sodium carbonate is given as 627 mg, which we can convert to grams:
627 e x t m g = 0.627 e x t g 627 ext{ mg} = 0.627 ext{ g} 627 e x t m g = 0.627 e x t g
To calculate the moles of Na2CO3, we need its molar mass. The formula for Na2CO3.xH2O can be written as:
e x t M o l a r m a s s = ( 2 i m e s 23 ) + 12 + ( 3 i m e s 16 ) + ( x i m e s 18 ) = 46 + 12 + 48 + 18 x = 60 + 18 x ext{Molar mass} = (2 imes 23) + 12 + (3 imes 16) + (x imes 18) = 46 + 12 + 48 + 18x = 60 + 18x e x t M o l a r ma ss = ( 2 im es 23 ) + 12 + ( 3 im es 16 ) + ( x im es 18 ) = 46 + 12 + 48 + 18 x = 60 + 18 x
Moles of Na2CO3 reacted Only available for registered users.
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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
Solving for x Only available for registered users.
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Cross-multiplying to solve for x:
0.627 = 0.00798 ( 60 + 18 x ) 0.627 = 0.00798(60 + 18x) 0.627 = 0.00798 ( 60 + 18 x )
This results in:
0.627 = 0.4788 + 0.14364 x 0.627 = 0.4788 + 0.14364x 0.627 = 0.4788 + 0.14364 x
Isolating x gives:
0.627 − 0.4788 = 0.14364 x 0.627 - 0.4788 = 0.14364x 0.627 − 0.4788 = 0.14364 x
0.1482 = 0.14364 x 0.1482 = 0.14364x 0.1482 = 0.14364 x
Thus,
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