n a titration experiment, 20.0 cm³ of aqueous sodium hydroxide reacts exactly with 25.0 cm³ of 0.100 mol/dm dilute sulfuric acid to make sodium sulfate.NaOH(aq) + H_{2}*S*O_{4}(aq) -> N*a_{2}*S*O_{4}(aq) + 2H_{3}*O(l)Some salts can be made by titration.(i) Circle the name of the type of reaction that takes place.[1]reductionprecipitation(ii) Calculate the concentration of the aqueous sodium hydroxide in g/dm³ using the following steps.Calculate the number of moles of dilute sulfuric acid used.mol5.decompositionneutralisation
Question
n a titration experiment, 20.0 cm³ of aqueous sodium hydroxide reacts exactly with 25.0 cm³ of 0.100 mol/dm dilute sulfuric acid to make sodium sulfate.NaOH(aq) + H_{2}SO_{4}(aq) -> N*a_{2}SO_{4}(aq) + 2H_{3}*O(l)Some salts can be made by titration.(i) Circle the name of the type of reaction that takes place.[1]reductionprecipitation(ii) Calculate the concentration of the aqueous sodium hydroxide in g/dm³ using the following steps.Calculate the number of moles of dilute sulfuric acid used.mol5.decompositionneutralisation
Solution
(i) The type of reaction that takes place is neutralisation.
(ii) To calculate the concentration of the aqueous sodium hydroxide, we first need to calculate the number of moles of dilute sulfuric acid used.
The number of moles of a substance can be calculated using the formula:
Moles = Volume (in dm³) x Concentration (in mol/dm³)
The volume of the sulfuric acid used in the experiment is given as 25.0 cm³. To convert this to dm³, we divide by 1000:
25.0 cm³ = 25.0 / 1000 = 0.025 dm³
The concentration of the sulfuric acid is given as 0.100 mol/dm³.
So, the number of moles of sulfuric acid is:
Moles = 0.025 dm³ x 0.100 mol/dm³ = 0.0025 mol
The balanced chemical equation shows that one mole of sulfuric acid reacts with two moles of sodium hydroxide. Therefore, the number of moles of sodium hydroxide used is twice the number of moles of sulfuric acid:
Moles of NaOH = 2 x 0.0025 mol = 0.005 mol
The volume of sodium hydroxide used is 20.0 cm³, or 0.020 dm³.
So, the concentration of the sodium hydroxide is:
Concentration = Moles / Volume = 0.005 mol / 0.020 dm³ = 0.25 mol/dm³
To convert this to g/dm³, we multiply by the molar mass of sodium hydroxide (NaOH), which is approximately 40 g/mol:
Concentration = 0.25 mol/dm³ x 40 g/mol = 10 g/dm³
So, the concentration of the aqueous sodium hydroxide is 10 g/dm³.
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