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Standard solutions of sodium hydroxide, NaOH, must be kept in airtight containers - VCE - SSCE Chemistry - Question 9 - 2016 - Paper 1

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Standard solutions of sodium hydroxide, NaOH, must be kept in airtight containers. This is because NaOH is a strong base and absorbs acidic oxides, such as carbon di... show full transcript

Worked Solution & Example Answer:Standard solutions of sodium hydroxide, NaOH, must be kept in airtight containers - VCE - SSCE Chemistry - Question 9 - 2016 - Paper 1

Step 1

a.i. Write a balanced overall equation for the reaction between CO₂ gas and water to form H₂CO₃.

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Answer

The balanced overall equation for the reaction between carbon dioxide and water is:

CO2(g)+H2O(l)H2CO3(aq)CO₂(g) + H₂O(l) \rightarrow H₂CO₃(aq)

Step 2

a.ii. Write a balanced equation for the complete reaction between H₂CO₃ and NaOH to form Na₂CO₃.

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Answer

The balanced equation for the complete reaction between carbonic acid and sodium hydroxide is:

H2CO3(aq)+2NaOH(aq)Na2CO3(aq)+2H2O(l)H₂CO₃(aq) + 2NaOH(aq) \rightarrow Na₂CO₃(aq) + 2H₂O(l)

Step 3

b.i. Calculate the amount of CO₂ in mol, that entered the container.

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Answer

To calculate the amount of CO₂ that entered the container, we can use the ideal gas law and concentration:

  1. Calculate the volume of air that entered the container, which is the volume of the container not filled with the NaOH solution: Vair=10.00 L9.90 L=0.10 LV_{air} = 10.00 \text{ L} - 9.90 \text{ L} = 0.10 \text{ L}

  2. Calculate the moles of CO₂ in the air: n(CO2)=extconcentration×Vairn(CO2)=0.0400%×0.10 L1extmol100extLn(CO₂) = ext{concentration} \times V_{air} \\ n(CO₂) = 0.0400 \% \times 0.10 \text{ L} \frac{1 ext{ mol}}{100 ext{ L}} =4.00×104 mol= 4.00 \times 10^{-4} \text{ mol}

  3. The conditions of the air can also be used with the ideal gas equation: Given P = 101.3 kPa, V = 0.10 L, R = 8.31 L·kPa/(K·mol), T = 21.5 °C = 294.5 K, n=PVRT=101.3×0.108.31×294.5=0.00400extmoln = \frac{PV}{RT} = \frac{101.3 \times 0.10}{8.31 \times 294.5} = 0.00400 ext{ mol}

Thus, the amount of CO₂ that entered the container is approximately 0.00040 mol.

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