Chapter 10, Problem 10.113QA

Interpretation Introduction

To find:

Henry’s law constant (${\mathrm{k}}_{\mathrm{H}})$ in $\frac{mol}{L\times atm}$ for O₂ dissolved in blood at $37℃$.

Answer to Problem 10.113QA

Solution:

The value for the Henry’s law constant for dissolved O₂ in arterial blood is

$7.8\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-3}\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{ }\times \mathrm{a}\mathrm{t}\mathrm{m}}$

Explanation of Solution

1) Concept:

Henry’s law states that, the solubility of any gas in a liquid is directly proportional to its partial pressure of the gas above the surface of the liquid.

${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}={\mathrm{k}}_{\mathrm{H}}\mathrm{ }\times \mathrm{ }{\mathrm{P}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$

Where, ${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$ is the solubility of gas in a particular solvent in mol/L, ${\mathrm{k}}_{\mathrm{H}}$ is the Henry’s gas constant, ${\mathrm{P}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$ is the partial pressure of gas in the aqueous solution in atm.

2) Given:

Arterial blood contains 0.25g O₂/L

3) Calculations:

${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}={\mathrm{k}}_{\mathrm{H}}\times {\mathrm{P}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$

The ${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$ has the unit mol/L. So, we must first find the molarity of O₂ in blood. Using molar mass of O₂, we calculate moles of O₂ and then molarity as follows:

$0.25\mathrm{ }\mathrm{g}\mathrm{ }{\mathrm{O}}_2\mathrm{ }\times \mathrm{ }\mathrm{ }\frac{1\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }{\mathrm{O}}_2}{31.998\mathrm{ }\mathrm{g}\mathrm{ }{\mathrm{O}}_2}=\mathrm{ }\mathrm{ }0.00781298\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }{\mathrm{O}}_2$

$\mathrm{M}\mathrm{o}\mathrm{l}\mathrm{a}\mathrm{r}\mathrm{i}\mathrm{t}\mathrm{y}=\frac{\left(\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }{\mathrm{O}}_2\right)}{\mathrm{L}\mathrm{ }\mathrm{o}\mathrm{f}\mathrm{ }\mathrm{b}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{d}}\mathrm{ }=\frac{\left(0.00781298\mathrm{ }\mathrm{m}\mathrm{o}\mathrm{l}\mathrm{ }{\mathrm{O}}_2\right)}{1\mathrm{ }\mathrm{L}\mathrm{ }\mathrm{b}\mathrm{l}\mathrm{o}\mathrm{o}\mathrm{d}}=\mathrm{ }0.00781298\mathrm{ }\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}}$

Inserting the values of ${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$ and $P_{gas}$ = 1 atm, we get the Henry’s law constant for O₂ as follows:

${\mathrm{C}}_{\mathrm{g}\mathrm{a}\mathrm{s}}={\mathrm{k}}_{\mathrm{H}}\times {\mathrm{P}}_{\mathrm{g}\mathrm{a}\mathrm{s}}$

$0.00781298\mathrm{ }\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}}={\mathrm{k}}_{\mathrm{H}}\mathrm{ }\times 1\mathrm{ }\mathrm{a}\mathrm{t}\mathrm{m}$

${\mathrm{k}}_{\mathrm{H}}=0.00781298\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{ }\times \mathrm{a}\mathrm{t}\mathrm{m}}$

(A mole is the SI unit of amount chemical substance. When writing units, it is written as “mol”.)

Conclusion:

The value for the Henry’s law constant for dissolved O₂ in arterial blood is

$7.8\mathrm{ }\mathrm{x}\mathrm{ }{10}^{-3}\frac{\mathrm{m}\mathrm{o}\mathrm{l}}{\mathrm{L}\mathrm{ }\times \mathrm{a}\mathrm{t}\mathrm{m}}$