Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha

Question

Gwar

2 months ago

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Arbon dioxide is dissolved in blood (ph 7.5) to form a mixture of carbonic acid and bicarbonate. Part a neglecting free co2, wha

t fraction will be present as carbonic acid? (pka for h2co3 and hco3− are 6.3 and 10.25, respectively)

Chemistry

2 answers:

castortr0y [3K] · Answer 1 · 2026-03-10T04:33:28+03:00

The proportion of carbonic acid present in the bloodstream is $\boxed{0.0595}$ .

Detailed Explanation:

Buffer solutions:

A buffer solution consists of either a weak acid along with its corresponding conjugate base or a weak base with its conjugate acid, which helps maintain stable pH levels when small quantities of acids or bases are added.

The mixture mentioned consists of carbonic acid in combination with bicarbonate, qualifying it as a buffer formed by a weak acid and its conjugate counterpart.

The pH for buffer solutions can be calculated using the Henderson-Hasselbalch equation, represented mathematically for this specific buffer as follows:

${\text{pH}} = {\text{p}}{K_{\text{a}}} + {\text{log}}\dfrac{{\left[ {{\text{HCO}}_3^ - } \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]}}$ …… (1)

Rearranging equation (1) produces:

$\log \dfrac{{\left[ {{\text{HCO}}_3^ - } \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]}} = {\text{pH}} - {\text{p}}{K_{\text{a}}}$ …… (2)

Given the pH is 7.5 and the value of ${\text{p}}{K_{\text{a}}}$ is 6.3.

Inserting these values into equation (2) yields:

$\begin{aligned}\log \frac{{\left[ {{\text{HCO}}_3^ - } \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]}}&= {\text{7}}{\text{.5}} - 6.3\\&= 1.2\\\end{aligned}$

Calculating the ratio of $\left[ {{\text{HCO}}_3^ - } \right]$ to $\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]$ , results in:

$\begin{aligned}\dfrac{{\left[ {{\text{HCO}}_3^ - }\right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}}\right]}} &= {10^{1.2}}\\&= 15.8\\\end{aligned}$

This indicates that the concentration of ${\text{HCO}}_3^ -$ is located at 15.8 times that of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ . Therefore, it's expressed as follows:

$\left[ {{\text{HCO}}_3^ - } \right]= 15.8{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]$ …… (3)

The cumulative mole fraction of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${\text{HCO}}_3^ -$ equals 1, leading to the following representation:

$\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] + \left[ {{\text{HCO}}_3^ - } \right] = 1$ …… (4)

By substituting equation (3) back into equation (4):

$\begin{aligned}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] + 15.8{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]&= 1 \hfill\\16.8{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] &= 1 \hfill\\\end{aligned}$

Now, solving for the concentration of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ ,

$\begin{aligned}\left[{{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]&= \frac{1}{{16.8}}\\&= 0.0595\\\end{aligned}$

Thus, the carbonic acid fraction in the bloodstream is 0.0595.

Additional Insights:

Water's acidity is explained here
The acidic and basic characteristics of amino acids are explored here

Response Information:

Grade: High School

Chapter: Acid, base and salts.

Subject: Chemistry

Keywords: pH, buffer, 6.3, 7.5, 1.2, 0.0595, 15.8, weak acid, conjugate base, H2CO3, HCO3-.

VMariaS [2.9K] · Answer 2 · 2026-03-10T04:32:25+03:00

Answer: The proportion of carbonic acid in the blood is 5.95%

Explanation:

The solution consists of carbonic acid (H₂CO₃) and bicarbonate ions (HCO₃⁻). This is a combination of a weak acid and its conjugate, forming a buffer.

The pH of a buffer can be determined using the Henderson equation, outlined below.

$pH = pKa + log \frac{[Base]}{[Acid]}$