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2 years ago

Neglecting free co2, what fraction will be present as carbonic acid? (pka for h2co3 and hco3− are 6.3 and 10.25, respectively)

Chemistry

2 answers:

notsponge [240]2 years ago

5 0

The correct answer is 5.93 %.

In blood, CO₂ dissolves and follow the following equilibrium:

CO₂ + H₂O ⇔ HCO₃⁻ + H⁺

In order to find the fraction of carbonic acid, we have to use Handerson-Hasselbalch equation, This equation illustrates the derivation of pH as a measure of acidity.

The equation is:

pH = PKa + log10 [A-] / [HA]

[HA] is the molar concentration of undissociated acid, and [A-] is the molar concentration of conjugate base.

Now, pH of blood is 7.5

PKa of H₂CO₃ is 6.3, and PKa for HCO₃⁻ is 10.25

Using equation,

pH = PKa + log10 [A-] / [HA]

Here, [A-] = [HCO₃⁻] and [HA] = [H₂CO₃]

7.5 = 6.3 + log10 [HCO₃]⁻ / [H₂CO₃]

7.5 - 6.3 = log10[HCO₃]⁻ / [H₂CO₃]

1.2 = log10[HCO₃]⁻ / [H₂CO₃]

10^(1.2) = [HCO₃]⁻ / [H₂CO₃]

15.84 = [HCO₃]⁻ / [H₂CO₃]

15.84 [H₂CO₃] = [HCO₃]⁻

Now concentration of [H₂CO₃] and its conjugate base must be equivalent to 1,

Therefore, [H₂CO₃] + [HCO₃]⁻ = 1

[H₂CO₃] + 15.84 [H₂CO₃] = 1

[H₂CO₃] [1 + 15.84] = 1

[H₂CO₃] [16.84] = 1

[H₂CO₃] = 1 / 16.84

[H₂CO₃] = 0.0593 = 5.93 %

san4es73 [151]2 years ago

3 0

The fraction of carbonic acid in the blood is $\boxed{0.00735}$ .

Further Explanation:

Buffer solution:

An aqueous solution of a weak acid and its conjugate base or a weak base and its conjugate acid is known as buffer solutions. When small amounts of acid or base are added to these solutions, these resist any change in their pH.

The given mixture contains carbonic acid and bicarbonate so it a buffer of a weak acid and its conjugate base. The dissociation of carbonic acid occurs as follows:

${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}} \rightleftharpoons {{\text{H}}^ + } + {\text{HCO}}_3^ -$

Henderson-Hasselbalch equation is used to determine the pH of buffer solutions. The mathematical form of this equation for the given buffer is 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)

Rearrange equation (1) to calculate the ratio of $\left[ {{\text{HCO}}_3^ - } \right]$ to $\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]$ .

$\dfrac{{\left[ {{\text{HCO}}_3^ - } \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]}}= {10^{{\text{pH}} - {\text{p}}{K_{\text{a}}}}}$ …… (2)

Substitute 7.4 for pH and 6.3 for ${\text{p}}{K_{\text{a}}}$ in equation (2).

$\begin{aligned}\frac{{\left[ {{\text{HCO}}_3^ - } \right]}}{{\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]}} &= {10^{7.4 - 6.3}}\\&= {10^{1.1}}\\&= 12.59\\\end{aligned}$

Therefore the concentration of ${\text{HCO}}_3^ -$ can be written in terms of that of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ by the following expression:

$\left[ {{\text{HCO}}_3^ - } \right]= 12.59{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right]$ …… (3)

The total mole fraction of ${{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}$ and ${\text{HCO}}_3^ -$ is 1. So it can be written as follows:

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

Substitute equation (3) in equation (4).

$\begin{aligned}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] + 12.59{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] &= 1 \hfill \\13.59{\text{ }}\left[ {{{\text{H}}_{\text{2}}}{\text{C}}{{\text{O}}_{\text{3}}}} \right] &= 1 \hfill\\\end{aligned}$