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weeeeeb [17]
3 years ago
15

Which one would be expected to happen? If one of the buffers that contribute to pH stability in human blood is carbonic acid (H2

CO3). Carbonic acid is a weak acid that when placed in an aqueous solution dissociates into a bicarbonate ion (HCO3-) and a hydrogen ion (H+). Thus, H2CO3 HCO3- + H+ If the pH of the blood increases.
(A) A decrease in the concentration of HC03- and an increase in the concentration of both H2CO3 and H2O
(B) An increase in the concentration of H2CO3 and a decrease in the concentration of H2O
(C) An increase in the concentration of HCO3- and a decrease in the concentration of H2O
(D) A decrease in the concentration of H2CO3 and an increase in the concentration of H2O
(E) A decrease in the concentration of HCO3- and an increase in the concentration of H2O
Chemistry
1 answer:
tino4ka555 [31]3 years ago
5 0

Answer:

Option (D) is definitely the answer.

Explanation:

Before going further, it is important to know what buffers and pH represent, which are keywords to answering this question.

Buffers is a special solution that can withstand or resist changes due to pH levels which may be as a result of an introduction of acidic or basic components into the blood. In other words, they maintain the stability of pH level in the human blood.

pH blood levels on the other hand, can be grouped into three: acidity, neutrality and alkalinity. Using a pH scale, one can determine its current level. In the human blood the pH level is near neutral and needs to be on a level near 7.4 in order to avoid a high rise or a drastic fall even if acidic or basic components come in or departs the blood stream.

Therefore, if one of the buffers that contributes to pH stability in human blood is carbonic acid, which is as a result of a combination of carbon dioxide and water in the blood stream. On getting to the lungs it is converted to water and subsequently released as waste. Maintaining this stability will definitely be to decrease the concentration of carbonic acid and increase that of water instead.  

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ln \frac{k_{2}}{k_{1}} = \frac{-E_{a}}{R}[\frac{1}{T_{2}} - \frac{1}{T_{1}}]

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