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Sodium thiosulfate is a reducing agent that reacts with iodine according to the following balanced equation - Leaving Cert Chemistry - Question 1 - 2015
Question 1
Sodium thiosulfate is a reducing agent that reacts with iodine according to the following balanced equation. \[ I_2 + 2S_2O_3^{2-} \rightarrow 2I^- + S_4O_6^{2-} \]... show full transcript
Worked Solution & Example Answer:Sodium thiosulfate is a reducing agent that reacts with iodine according to the following balanced equation - Leaving Cert Chemistry - Question 1 - 2015
Step 1
Explain how iodine, a non-polar substance of very low water solubility, is brought into aqueous solution.
Answer
Iodine can be brought into an aqueous solution by using a non-polar solvent, such as chloroform or an organic solvent, to first dissolve the iodine. Once the iodine is dissolved, it can then be mixed with a suitable surfactant or emulsifier that allows it to interact with water, forming a stable emulsion. This allows some iodine to migrate into the aqueous phase, enabling it to be available for reaction in solution.
Step 2
Describe the procedure for measuring 25.0 cm³ of this solution into a conical flask.
Answer
To measure 25.0 cm³ of the iodine solution into a conical flask, first ensure the volumetric flask containing the iodine solution is properly mixed. Use a clean pipette, and carefully insert it into the flask, then draw the iodine solution into the pipette while keeping it vertical. When the meniscus aligns with the 25.0 cm³ mark, remove the pipette. Quickly transfer the pipette contents to the conical flask, ensuring to touch the tip of the pipette against the inside of the flask to minimize any loss of solution.
Step 3
Step 4
Step 5
Given that there were 6.35 g of iodine (I2) in 500 cm³ of the iodine solution, calculate the number of moles of iodine in each 25.0 cm³ portion.
Answer
To calculate the number of moles of iodine in each 25.0 cm³ portion: Molar mass of I2 = 253.8 g/mol. Total moles in 500 cm³: [ \text{Moles} = \frac{6.35 g}{253.8 g/mol} = 0.0250 \text{ moles} ] Thus, moles in 25.0 cm³: [ 0.0250 \text{ moles} \times \frac{25.0 cm³}{500 cm³} = 0.00125 \text{ moles} ]
Step 6
The number of moles of sodium thiosulfate required to reduce this quantity of iodine.
Answer
For the reaction between iodine and sodium thiosulfate, the stoichiometry shows that 1 mole of I2 reacts with 2 moles of Na2S2O3. Therefore, the moles of sodium thiosulfate required for 0.00125 moles of iodine is: [ 0.00125 \text{ moles I2} \times 2 \frac{moles Na2S2O3}{mole I2} = 0.00250 \text{ moles Na2S2O3} ]
Step 7
The concentration of the sodium thiosulfate solution in moles per litre.
Answer
Given that the sodium thiosulfate solution is required to react with 25.0 cm³ of iodine, we can calculate its concentration: [ C = \frac{n}{V} = \frac{0.00250 \text{ moles}}{0.0250 \text{ L}} = 0.100 \text{ mol/L} ]
Step 8
The concentration of the sodium thiosulfate solution in grams per litre of its crystals (Na2S2O3·5H2O).
Answer
The molar mass of Na2S2O3·5H2O is about 248.18 g/mol. Therefore, the concentration in grams per litre can be calculated as follows: [ \text{Concentration (g/L)} = 0.100 \text{ mol/L} \times 248.18 \text{ g/mol} = 24.8 \text{ g/L} ]
Step 9
Explain why the use of distilled water instead of deionized water throughout this experiment would be likely to ensure a more accurate result.
Answer
Using distilled water instead of deionized water can introduce contaminants that may affect the ionic concentration of the solutions involved. Deionized water is free from ions and impurities, ensuring that no additional variables influence the results. This maintains the integrity of the titration reaction and leads to more accurate measurements of the endpoint.