# 8.6 Buffers  (Page 5/9)

 Page 5 / 9

## Medicine: the buffer system in blood

The normal pH of human blood is about 7.4. The carbonate buffer system in the blood uses the following equilibrium reaction:

${\text{CO}}_{2}\left(g\right)+2{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{CO}}_{3}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{HCO}}_{3}{}^{\text{−}}\left(aq\right)+{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)$

The concentration of carbonic acid, H 2 CO 3 is approximately 0.0012 M , and the concentration of the hydrogen carbonate ion, ${\text{HCO}}_{3}{}^{\text{−}},$ is around 0.024 M . Using the Henderson-Hasselbalch equation and the p K a of carbonic acid at body temperature, we can calculate the pH of blood:

$\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\phantom{\rule{0.4em}{0ex}}\frac{\left[\text{base}\right]}{\left[\text{acid}\right]}\phantom{\rule{0.2em}{0ex}}=6.4+\text{log}\phantom{\rule{0.4em}{0ex}}\frac{0.024}{0.0012}\phantom{\rule{0.2em}{0ex}}=7.7$

The fact that the H 2 CO 3 concentration is significantly lower than that of the ${\text{HCO}}_{3}{}^{\text{−}}$ ion may seem unusual, but this imbalance is due to the fact that most of the by-products of our metabolism that enter our bloodstream are acidic. Therefore, there must be a larger proportion of base than acid, so that the capacity of the buffer will not be exceeded.

Lactic acid is produced in our muscles when we exercise. As the lactic acid enters the bloodstream, it is neutralized by the ${\text{HCO}}_{3}{}^{\text{−}}$ ion, producing H 2 CO 3 . An enzyme then accelerates the breakdown of the excess carbonic acid to carbon dioxide and water, which can be eliminated by breathing. In fact, in addition to the regulating effects of the carbonate buffering system on the pH of blood, the body uses breathing to regulate blood pH. If the pH of the blood decreases too far, an increase in breathing removes CO 2 from the blood through the lungs driving the equilibrium reaction such that [H 3 O + ] is lowered. If the blood is too alkaline, a lower breath rate increases CO 2 concentration in the blood, driving the equilibrium reaction the other way, increasing [H + ] and restoring an appropriate pH.

## Key concepts and summary

A solution containing a mixture of an acid and its conjugate base, or of a base and its conjugate acid, is called a buffer solution. Unlike in the case of an acid, base, or salt solution, the hydronium ion concentration of a buffer solution does not change greatly when a small amount of acid or base is added to the buffer solution. The base (or acid) in the buffer reacts with the added acid (or base).

## Key equations

• p K a = −log K a
• p K b = −log K b
• $\text{pH}=\text{p}{K}_{\text{a}}+\text{log}\phantom{\rule{0.2em}{0ex}}\frac{\left[{\text{A}}^{\text{−}}\right]}{\left[\text{HA}\right]}$

Explain why a buffer can be prepared from a mixture of NH 4 Cl and NaOH but not from NH 3 and NaOH.

Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the acid H 3 PO 4 and a salt of its conjugate base NaH 2 PO 4 .

Excess H 3 O + is removed primarily by the reaction:
${\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)$
Excess base is removed by the reaction:
${\text{OH}}^{\text{−}}\left(aq\right)+{\text{H}}_{3}{\text{PO}}_{4}\left(aq\right)\phantom{\rule{0.2em}{0ex}}⟶\phantom{\rule{0.2em}{0ex}}{\text{H}}_{2}{\text{PO}}_{4}{}^{\text{−}}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)$

Explain why the pH does not change significantly when a small amount of an acid or a base is added to a solution that contains equal amounts of the base NH 3 and a salt of its conjugate acid NH 4 Cl.

What is [H 3 O + ] in a solution of 0.25 M CH 3 CO 2 H and 0.030 M NaCH 3 CO 2 ?
${\text{CH}}_{3}{\text{CO}}_{2}\text{H}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{CH}}_{3}{\text{CO}}_{2}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}}=1.8\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

[H 3 O + ] = 1.5 $×$ 10 −4 M

What is [H 3 O + ] in a solution of 0.075 M HNO 2 and 0.030 M NaNO 2 ?
${\text{HNO}}_{2}\left(aq\right)+{\text{H}}_{2}\text{O}\left(l\right)\phantom{\rule{0.2em}{0ex}}⇌\phantom{\rule{0.2em}{0ex}}{\text{H}}_{3}{\text{O}}^{\text{+}}\left(aq\right)+{\text{NO}}_{2}{}^{\text{−}}\left(aq\right)\phantom{\rule{5em}{0ex}}{K}_{\text{a}}=4.5\phantom{\rule{0.2em}{0ex}}×\phantom{\rule{0.2em}{0ex}}{10}^{-5}$

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