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The Earth’s oceans contain significant amounts of dissolved carbon dioxide - VCE - SSCE Chemistry - Question 6 - 2003 - Paper 1

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The Earth’s oceans contain significant amounts of dissolved carbon dioxide. The dissolving process can be described by the following chemical equilibria. $$CO_2(g) ... show full transcript

Worked Solution & Example Answer:The Earth’s oceans contain significant amounts of dissolved carbon dioxide - VCE - SSCE Chemistry - Question 6 - 2003 - Paper 1

Step 1

Use this information to explain the likely effect of the increasing concentration of atmospheric CO2 on the pH of seawater at the ocean surface.

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Answer

Increasing the concentration of atmospheric CO2CO_2 leads to a shift in the first equilibrium to the right, which increases the amount of dissolved CO2(aq)CO_2(aq). According to Le Chatelier’s principle, the equilibrium will shift to counteract this change. This increased concentration of CO2(aq)CO_2(aq) will push the second equilibrium to the right, further increasing the hydrogen ion concentration, [H+][H^+]. Therefore, as [H+][H^+] increases, the pH of seawater will decrease, indicating increased acidity.

Step 2

i. Calculate the concentration of the hydrogen ion, H+, in fresh blood.

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Answer

Given that the pH of blood is 7.4, we can calculate the concentration of H+H^+ using the formula:

pH=−log⁡[H+]pH = -\log[H^+]

Rearranging this gives:

[H+]=10−pH=10−7.4≈3.98×10−8M[H^+] = 10^{-pH} = 10^{-7.4} \approx 3.98 \times 10^{-8} M.

Step 3

ii. Calculate the concentration of the hydrogen carbonate ion, HCO3-, in fresh blood.

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Answer

Using the provided equilibrium constant,

Ka=[H+][HCO3−][CO2]K_a = \frac{[H^+][HCO_3^-]}{[CO_2]}

we substitute the known values:

  • [H+]≈3.98×10−8M[H^+] \approx 3.98 \times 10^{-8} M
  • [CO2]=1.3×10−5M[CO_2] = 1.3 \times 10^{-5} M
  • Ka=7.9×10−7K_a = 7.9 \times 10^{-7}

Now, rearranging gives:

[HCO3−]=Ka⋅[CO2][H+]=(7.9×10−7)⋅(1.3×10−5)3.98×10−8≈2.59×10−2M.[HCO_3^-] = \frac{K_a \cdot [CO_2]}{[H^+]} = \frac{(7.9 \times 10^{-7}) \cdot (1.3 \times 10^{-5})}{3.98 \times 10^{-8}} \approx 2.59 \times 10^{-2} M.

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