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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. CO2(g) ⇌ C... 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

As the concentration of atmospheric CO2 increases, the amount of CO2 that dissolves in seawater also increases. This higher concentration of dissolved CO2 shifts the first equilibrium to the right, resulting in an increase in the concentration of CO2(aq). According to Le Chatelier's principle, this will lead to a rightward shift in the second equilibrium as well, producing more hydrogen ions (H+) and bicarbonate ions (HCO3−). The increase in hydrogen ions will result in a decrease in pH, making the seawater more acidic.

Step 2

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

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Answer

The pH of blood is given as 7.4. We can calculate the concentration of hydrogen ions using the formula:

[H+]=10−pH[H^+] = 10^{-pH}

Substituting the known pH value:

[H+]=10−7.4[H^+] = 10^{-7.4}

Calculating this gives:

[H+]≈3.98×10−8 M[H^+] \approx 3.98 \times 10^{-8} \text{ M}

Step 3

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

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Answer

Using the equilibrium constant expression,

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

Given that Ka=7.9×10−7K_a = 7.9 \times 10^{-7}, [H+]≈3.98×10−8 M[H^+] \approx 3.98 \times 10^{-8} \text{ M}, and [CO2]=1.3×10−5 M[CO2] = 1.3 \times 10^{-5} \text{ M}, we can rearrange the equation to solve for [HCO3−][HCO3^-]:

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

Substituting the known values:

[HCO3−]=(7.9×10−7)×(1.3×10−5)3.98×10−8[HCO3^-] = \frac{(7.9 \times 10^{-7}) \times (1.3 \times 10^{-5})}{3.98 \times 10^{-8}}

Calculating this gives:

[HCO3−]≈2.56×10−6 M[HCO3^-] \approx 2.56 \times 10^{-6} \text{ M}

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