# Question #4bb1b

Jan 30, 2015

The dominant form of the phosphoric acid is ${\text{HPO}}_{4}^{2 -}$ and its concentration is $\text{0.639 mmol/L}$.

${H}_{3} P {O}_{4} + {H}_{2} O r i g h t \le f t h a r p \infty n s {H}_{2} P {O}_{4}^{-} + {H}_{3}^{+} O$, $p K {a}_{1} = 2.12$
${H}_{2} P {O}_{4}^{-} + {H}_{2} O r i g h t \le f t h a r p \infty n s H P {O}_{4}^{2 -} + {H}_{3}^{+} O$, $p K {a}_{2} = 7.21$
$H P {O}_{4}^{2 -} + {H}_{2} O r i g h t \le f t h a r p \infty n s P {O}_{4}^{3 -} + {H}_{3}^{+} O$, $p K {a}_{3} = 12.7$

Because the pH of blood lies between $p K {a}_{2}$ and $p K {a}_{3}$, the dominant form of the acid will be $H P {O}_{4}^{2 -}$. Because $p K {a}_{2}$ and $p K {a}_{3}$ vary by more than 4 units, the concentration of $P {O}_{4}^{3 -}$ will be negligible.

So, you know that the total phosphate concentration is $\text{1.05 mmol/L}$. This means that

$\left[{H}_{2} P {O}_{4}^{-}\right] + \left[H P {O}_{4}^{2 -}\right] = 1.05 \text{mmol/L}$ (assuming $\left[P {O}_{4}^{3 -}\right]$ is negligible). (1)

Since you're dealing with a buffer, the Henderson-Hasselbalch equation can be used

$\left[p H\right] \left(h \texttt{p} : / s o c r a t i c . \mathmr{and} \frac{g}{c} h e m i s t r \frac{y}{a} c i \mathrm{ds} - \mathmr{and} - b a s e \frac{s}{t} h e - p h - c o n c e p t\right) = p K {a}_{2} + \log \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right)$

$7.40 = 7.21 + \log \left(\frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]}\right) \implies \frac{\left[H P {O}_{4}^{2 -}\right]}{\left[{H}_{2} P {O}_{4}^{-}\right]} = 1.55$

Plug this value into (1) and you'll get

$\frac{\left[H P {O}_{4}^{2 -}\right]}{1.55} + \left[H P {O}_{4}^{2 -}\right] = 1.05$, which will result in

$\left[H P {O}_{4}^{2 -}\right] = \text{0.639 mmol/L}$

The concentration of ${H}_{2} P {O}_{4}^{-}$ will be $\left[{H}_{2} P {O}_{4}^{-}\right] = \text{0.412 mmol/L}$

If you're interested, you can check the assumption that $\left[P {O}_{4}^{3 -}\right]$ is negligible; the equations will produce

$\left[P {O}_{4}^{3 -}\right] = \text{0.00000320 mmol/L}$ $\to$ for all intended purposes, this is equal to zero.

One more thing...the results match the known proportions of dyhidrogen phosphate and hydrogen phosphate in extracellular fluid (39% for the former, 61% for the latter).