Question

Phosphate buffers are important in regulating the \(\mathrm{pH}\) of intracellular fluids at \(\mathrm{pH}\) values generally between \(7.1\) and \(7.2\). What is the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15 ?\) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8}\) Why is a buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid? \(\mathrm{H}_{3} \mathrm{PO}_{4}(a q) \rightleftharpoons \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=7.5 \times 10^{-3}\)

Short answer

The concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}^{2-}\) in intracellular fluid at \(\mathrm{pH}=7.15\) is approximately \(0.595\). A buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is ineffective in buffering the pH of intracellular fluid because the pKa value of the first dissociation step is not close enough to the desired pH range, resulting in a buffering capacity that is more effective at a lower pH range than that of intracellular fluids.

Step-by-step solution

1. Apply Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a solution to the pKa and the concentration ratio of the weak acid and its conjugate base: \[ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{\text{conjugate base}}{\text{weak acid}}\right) \] Here, we want to find the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) (conjugate base) to \(\mathrm{HPO}_{4}^{2-}\) (weak acid).

2. Rearrange the Henderson-Hasselbalch equation to solve for the concentration ratio

In order to find the concentration ratio, we need to rearrange the equation as follows: \[ \frac{\text{conjugate base}}{\text{weak acid}} = 10^{\mathrm{pH} - \mathrm{p}K_a} \]

3. Plug in the given values and solve for the concentration ratio

We have the following values: \[\mathrm{pH} = 7.15\] \[K_a = 6.2 \times 10^{-8}\] To obtain the pKa value, we take the negative logarithm of the Ka value: \[ \mathrm{p}K_a = -\log(6.2 \times 10^{-8}) \] Now, plug in these values into the rearranged Henderson-Hasselbalch equation and calculate the concentration ratio: \[ \frac{\mathrm{H}_{2} \mathrm{PO}_{4}^{-}}{\mathrm{HPO}_{4}^{2-}} = 10^{7.15 - (-\log(6.2 \times 10^{-8}))} \] This gives us the concentration ratio: \[ \frac{\mathrm{H}_{2} \mathrm{PO}_{4}^{-}}{\mathrm{HPO}_{4}^{2-}} \approx 0.595 \]

4. Explain the ineffectiveness of the H3PO4 and H2PO4- buffer

A buffer composed of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) and \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is ineffective in buffering the pH of intracellular fluid because the Ka value for the dissociation of \(\mathrm{H}_{3} \mathrm{PO}_{4}\) is too large (\(K_a = 7.5 \times 10^{-3}\)), resulting in a relatively low pKa value. This translates to a buffering capacity that is more effective at a lower pH range than that of intracellular fluids, which have a pH range of around \(7.1\) to \(7.2\). The pKa value of the first dissociation step is not close enough to the desired pH range so it will not buffer efficiently at the desired pH.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a crucial tool in chemistry for understanding how the pH of a buffer solution is related to the concentrations of acid and base components present. It expresses the relationship between pH, the acid dissociation constant (pKa), and the concentration ratio of the conjugate base to the weak acid. The equation is given by: \[ \mathrm{pH} = \mathrm{p}K_a + \log\left(\frac{\text{conjugate base}}{\text{weak acid}}\right) \]This equation is particularly useful when you need to calculate the pH of a buffer system, or to determine the required ratio of concentrations of acid and base needed to achieve a desired pH. In the context of the phosphate buffer system, which is significant for maintaining pH levels in biological fluids, the equation allows us to balance the concentrations of the hydrogen phosphate and dihydrogen phosphate ions.
The beauty of the Henderson-Hasselbalch equation lies in its ability to offer insights using logarithms, simplifying computational work that can otherwise be cumbersome with direct concentration calculations.

Conjugate Acid-Base Pair

A conjugate acid-base pair consists of two species that transform into each other by the gain or loss of a proton (\(\mathrm{H}^+\)). In a phosphate buffer system, the conjugate acid-base pair involves \(\mathrm{H}_{2}\mathrm{PO}_4^-\) (dihydrogen phosphate, acting as the acid) and \(\mathrm{HPO}_4^{2-}\) (hydrogen phosphate, acting as the base).
This pair is effective in maintaining pH around neutral levels, as it can either donate or accept protons in response to changes in the environment.

• When the pH of the solution decreases, suggesting an excess of \(\mathrm{H}^+\) ions, the hydrogen phosphate ions will capture these protons, turning into dihydrogen phosphate ions.
• Conversely, if the pH increases, indicating a shortage of \(\mathrm{H}^+\) ions, dihydrogen phosphate ions donate protons, converting to hydrogen phosphate ions.

This back-and-forth transformation allows the system to resist significant pH changes, providing the stability crucial for biological processes.

Buffer Capacity

Buffer capacity refers to the ability of a buffer solution to resist changes in pH upon addition of an acid or base. The effectiveness of a buffer depends on the concentrations of its acid-base pair and the closeness of its pKa to the desired pH. In simpler terms, a buffer has a higher capacity if it can absorb more added acid or base without a significant change in pH.
The phosphate buffer system, with its components \(\mathrm{H}_{2}\mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\), is effective in physiological pH ranges (around 7.1 to 7.2) due to its pKa value aligning closely with this range. However, a buffer system involving \(\mathrm{H}_{3}\mathrm{PO}_4\) and \(\mathrm{H}_{2}\mathrm{PO}_4^-\) is less effective here because its pKa is more suited for buffering at a lower pH. This is why the exercise mentions its inefficiency; its capacity is not strong enough at the cellular pH levels.