Question

There is a buffer system \(\left(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}{ }^{2-}\right)\) in blood that helps keep the blood \(\mathrm{pH}\) at about \(7.40 .\left(\mathrm{K}_{\mathrm{a}} \mathrm{H}_{2} \mathrm{PO}_{4}^{-}=6.2 \times 10^{-8}\right)\). (a) Calculate the \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right] /\left[\mathrm{HPO}_{4}^{2-}\right]\) ratio at the normal \(\mathrm{pH}\) of blood. (b) What percentage of the \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when the \(\mathrm{pH}\) goes down to \(6.80\) ? (c) What percentage of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\) when the \(\mathrm{pH}\) goes up to \(7.80\) ?

Short answer

To summarize, the ratio \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]/\left[\mathrm{HPO}_{4}^{2-}\right]\) at the normal blood pH of 7.40 is 1.58. When the pH goes down to 6.80, 83.6% of \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\). When the pH goes up to 7.80, 149.5% of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\).

Step-by-step solution

Calculate the ratio of [H2PO4-]/[HPO42-] at pH 7.40

We need to find the ratio \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]/\left[\mathrm{HPO}_{4}^{2-}\right]\) at the normal blood pH of 7.40. We'll use the Henderson-Hasselbalch equation for this purpose: Henderson-Hasselbalch equation: \(pH = pK_{a} + log\left(\frac{[A^-]}{[HA]}\right)\), where \([A^-]\) represents \(\left[\mathrm{HPO}_{4}^{2-}\right]\) and \([HA]\) represents \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]\) At pH = 7.40 and \(K_a = 6.2 \times 10^{-8}\), first calculate \(pK_a\): \(pK_a = -log(K_a) = -log(6.2 \times 10^{-8}) = 7.21\) Now, rearrange the equation for the ratio: \(\frac{[A^-]}{[HA]} = 10^{(pH - pK_a)} = 10^{(7.40 - 7.21)}\) Calculate the resulting ratio: \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]/\left[\mathrm{HPO}_{4}^{2-}\right] = 10^{(7.40 - 7.21)} = 1.58\)

Calculate percentage conversion at pH 6.80

Determine the conversion of \(\mathrm{HPO}_{4}{ }^{2-}\) to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) when pH decreases to 6.80. We'll use the same Henderson-Hasselbalch equation as in the previous step: Find the new ratio when pH = 6.80: \(\frac{[A^-]}{[HA]} = 10^{(pH - pK_a)} = 10^{(6.80 - 7.21)} = 0.258\) Now we know that the new ratio is 0.258. To calculate the percentage conversion, we can compare the ratio at pH 6.80 with that at pH 7.40: \(\frac{\text{New ratio - Original ratio}}{\text{Original ratio}} = \frac{0.258 - 1.58}{1.58} = -0.836\) Percentage conversion = -0.836 * 100% = -83.6% The negative sign indicates that at pH 6.80, 83.6% of \(\mathrm{HPO}_{4}{ }^{2-}\) ions are converted to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\).

Calculate percentage conversion at pH 7.80

Determine the conversion of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) to \(\mathrm{HPO}_{4}{ }^{2-}\) when pH increases to 7.80. We'll use the same Henderson-Hasselbalch equation as in the previous steps: Find the new ratio when pH = 7.80: \(\frac{[A^-]}{[HA]} = 10^{(pH - pK_a)} = 10^{(7.80 - 7.21)} = 3.944\) Now we know that the new ratio is 3.944. To calculate the percentage conversion, we can compare the ratio at pH 7.80 with that at pH 7.40: \(\frac{\text{New ratio - Original ratio}}{\text{Original ratio}} = \frac{3.944 - 1.58}{1.58} = 1.495\) Percentage conversion = 1.495 * 100% = 149.5% At pH 7.80, 149.5% of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ions are converted to \(\mathrm{HPO}_{4}{ }^{2-}\).

Key concepts

Buffer System

The human body meticulously regulates its pH levels, where slight deviations can disrupt normal functions. A buffer system plays a critical role in maintaining this stability. A buffer is a mixture of a weak acid and its conjugate base (or a weak base and its conjugate acid) that can resist changes in pH upon the addition of small amounts of acid or base.

In the exercise, the buffer system in focus is \( \left(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}^{2-}\right) \), which is based on dihydrogen phosphate ions (the weak acid, \( \mathrm{H}_{2}\mathrm{PO}_{4}^{-} \) and hydrogen phosphate ions (the conjugate base, \( \mathrm{HPO}_{4}^{2-} \)). When acid or base is added to the blood, these components react by either releasing or absorbing free hydrogen ions (\( H^+ \)), thus minimizing any drastic shift in pH.

Blood pH Regulation

Blood pH regulation is vital for bodily functions. The normal pH range of blood is very narrow, approximately 7.35 to 7.45, and the body uses several mechanisms to keep it within this range. The buffer system, mentioned earlier, is one of these mechanisms.

If the pH of the blood falls outside of this range, it can lead to acidosis (if too acidic) or alkalosis (if too alkaline), both of which can be life-threatening. Ensuring that the concentration of the buffer components (the acid and the base) is appropriate, is essential for the regulation of acid-base balance in the blood. The Henderson-Hasselbalch equation is a mathematical representation that helps in quantifying this balance, informing us of the concentration ratios of the buffer components necessary to maintain a specific pH.

Acid-Base Equilibrium

Acid-base equilibrium refers to the state of balance between acids and bases in a solution, which in turn, affects the pH of that solution. This equilibrium is governed by the principles of chemical equilibria, acting at the molecular level.

The Henderson-Hasselbalch equation is derived from the acid dissociation constant (\( K_a \)), which quantifies the strength of an acid in a solution. In essence, this equation allows us to calculate the pH of a buffer solution by using the concentrations of the acid and its conjugate base. This calculation thereby makes it easier to understand how different concentrations would affect the overall pH—a key concept that guides the pH adjustment in medical, biological, and chemical applications.

In the exercise provided, this concept is applied to determine how the blood buffer system responds to changes in pH, through various percentage conversion calculations.