Question

One of the buffer systems used to control the pH of blood involves the equilibrium between \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). The \(\mathrm{p} K_{\mathrm{a}}\) for \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) is 7.21 . (a) Write the Henderson-Hasselbalch equation for this buffer system. (b) What \(\mathrm{HPO}_{4}^{2-}\) to \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) ratio is needed to maintain the optimum blood \(\mathrm{pH}\) of \(7.40 ?\)

Short answer

The ratio \(\frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]}\) is approximately 1.55.

Step-by-step solution

Write the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation relates the pH, pKa, and the ratio of the concentrations of the base (\( ext{A}^-\)) to the acid (\( ext{HA}\)) in a buffer solution. For this buffer system involving \( ext{H}_2 ext{PO}_4^-\) and \( ext{HPO}_4^{2-}\), the equation is written as:\[\text{pH} = ext{p}K_a + ext{log} \left( \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \right)\]Given \( ext{p}K_a = 7.21\), substituting it into the equation gives:\[\text{pH} = 7.21 + ext{log} \left( \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \right)\]

Set the Equation for Desired pH

We need to calculate the ratio \(\frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]}\) that would give the blood an optimum pH of 7.40. Set \( ext{pH} = 7.40\) and substitute into the Henderson-Hasselbalch equation:\[7.40 = 7.21 + ext{log} \left( \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \right)\]

Solve for the Ratio

Subtract 7.21 from both sides to isolate the logarithmic term:\[7.40 - 7.21 = ext{log} \left( \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \right)\]\[0.19 = ext{log} \left( \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \right)\]To remove the logarithm, raise 10 to the power of both sides:\[10^{0.19} = \frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]}\]Calculate \(10^{0.19} \approx 1.55\). Thus, the ratio needed is:\[\frac{[ ext{HPO}_4^{2-}]}{[ ext{H}_2 ext{PO}_4^-]} \approx 1.55\]

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a vital tool for understanding buffer systems in chemistry. It relates the pH of a solution to the pKa and the ratio of the concentrations of an acid and its conjugate base. The equation is given by:

• \[\text{pH} = \text{pK}_a + \text{log} \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\]

This equation helps us calculate the pH of buffer solutions by knowing the pKa value of the acid and the concentrations of the acid and base. In the case of the phosphate buffer system involving \(\mathrm{H}_2 \mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\), the equation plays a crucial role in maintaining the desired blood pH levels. By understanding how to use the Henderson-Hasselbalch equation, you can predict how a buffer will respond to changes in concentration, providing insights into controlling the pH of various solutions.

blood pH

Blood pH is a critical measure of the oxygen and carbon dioxide balance in the bloodstream. It is typically maintained around 7.40, which is slightly alkaline. Deviations from this range can significantly impact bodily functions as enzymes and biochemical processes are sensitive to pH changes. Maintaining this balance is essential for the body's homeostasis. Our body utilizes several buffer systems to manage blood pH. These systems help resist drastic changes when acids or bases are introduced. Buffer systems, like the phosphate buffer system, play an essential role by reacting quickly with added acids or bases to keep the pH within the narrow, safe range. An effective buffer minimizes the pH shift despite small disturbances, ensuring that physiological processes occur efficiently and without disruption.

acid-base equilibrium

The concept of acid-base equilibrium is fundamental in understanding how buffers function. This equilibrium involves acids producing hydrogen ions (H+) and bases accepting them. The balance between these ions determines the pH of the solution.In buffer solutions, an equilibrium exists between a weak acid and its conjugate base or a weak base and its conjugate acid. The presence of a buffer means the solution can neutralize added acids or bases, thus maintaining a steady pH. For the phosphate buffer system, this involves the equilibrium between \(\mathrm{H}_2 \mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\). Adjusting the concentration of these components shifts the equilibrium to stabilize the pH.Understanding acid-base equilibrium is essential for predicting reactions in biological systems and in the design of chemical processes where pH control is crucial.

phosphate buffer system

The phosphate buffer system is particularly important in biological contexts for controlling pH within cells and biological fluids. This system includes the dihydrogen phosphate ion \(\mathrm{H}_2 \mathrm{PO}_4^-\) and hydrogen phosphate ion \(\mathrm{HPO}_4^{2-}\). It is most effective at a physiological pH level around 7.40, which is critical for many cellular processes.The phosphate buffer system operates by adjusting the balance between \(\mathrm{H}_2 \mathrm{PO}_4^-\) (acting as an acid) and \(\mathrm{HPO}_4^{2-}\) (acting as a base). When the pH increases, the system shifts to produce more \(\mathrm{H}_2 \mathrm{PO}_4^-\), and when the pH decreases, it shifts to produce more \(\mathrm{HPO}_4^{2-}\). This dynamic adjustment helps maintain a stable pH environment, which is critical for enzyme function, nutrient absorption, and overall cell health.By understanding how the phosphate buffer system works, you can appreciate its importance in both chemistry and biology, especially in maintaining the delicate balance necessary for life processes.