Question

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

a. The concentration ratio of \(\mathrm{H}_2\mathrm{PO}_4^-\) to \(\mathrm{HPO}_4^{2-}\) in intracellular fluid at \(\mathrm{pH} = 7.15\) is: \[\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]} = 10^{(7.15 - pK_a)}\] b. The buffer composed of \(\mathrm{H}_3\mathrm{PO}_4\) and \(\mathrm{H}_2\mathrm{PO}_4^-\) is ineffective in buffering the \(\mathrm{pH}\) of intracellular fluid because the \(pK_a\) value for the \(H_3PO_4\) reaction is significantly larger than the desired pH range, making it a poor buffer for maintaining the pH of intracellular fluid. Effective buffers have \(pK_a\) values close to the pH range to be maintained.

Step-by-step solution

The Henderson-Hasselbalch equation relates pH, the acid dissociation constant (\(K_a\)), and the concentrations of the acid and its conjugate base in a buffer solution. For this problem, the equation is given as: \[pH = pK_a + log \frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]}\] #Step 2: Find \(pK_a\) from given \(K_a\) values#

To calculate the concentration ratio, first, we need to find the \(pK_a\) values for the given dissociation constants \(K_a\). For \(H_2PO_4^- \rightleftharpoons HPO_4^{2-} + H^+\), \(K_a = 6.2 \times 10^{-8}\), so: \[pK_a = -log(6.2 \times 10^{-8})\] Similarly, for \(H_3PO_4 \rightleftharpoons H_2PO_4^- + H^+\), \(K_a = 7.5 \times 10^{-3}\), so: \[pK_a = -log(7.5 \times 10^{-3})\] #Step 3: Calculate the concentration ratio of \(H_2PO_4^-\) to \(HPO_4^{2-}\) at pH 7.15#

Here, the pH of intracellular fluids is given as 7.15, and we know the \(pK_a\) value calculated in step 2. We will now use the Henderson-Hasselbalch equation to find the concentration ratio: \[7.15 = pK_a + log \frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]}\] Solving for the ratio: \[\frac{[\mathrm{HPO}_4^{2-}]}{[\mathrm{H}_2\mathrm{PO}_4^-]} = 10^{(7.15 - pK_a)}\] #Step 4: Explain why the buffer composed of \(H_3PO_4\) and \(H_2PO_4^-\) is ineffective#

The buffer composed of \(H_3PO_4\) and \(H_2PO_4^-\) would be ideal if the pK_a value were close to the pH at which we want to maintain the intracellular fluid (7.1-7.2). However, the calculated pK_a value for the \(H_3PO_4\) reaction (Step 2) is significantly larger than the desired pH range, making it a poor buffer in maintaining the pH of intracellular fluid. Effective buffers have pK_a values close to the pH range to be maintained.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a valuable tool in chemistry for understanding how buffer solutions work in maintaining a stable pH in various environments, such as biological systems. It relates the pH of a solution with the pKa, which is the negative logarithm of the acid dissociation constant (Ka), and the concentrations of the acid and its conjugate base.
The equation is expressed as:

\[pH = pK_a + \log \frac{[\mathrm{base}]}{[\mathrm{acid}]}\]

In this formula, the 'base' is the conjugate base (the species the acid turns into after losing a hydrogen ion), and the 'acid' is the original form of the substance able to donate a hydrogen ion. This equation helps predict the pH of a buffer solution by showing how changes in the ratio of these components affect the pH.

For example, in a phosphate buffer system, we apply this equation to find out the concentration ratio needed to maintain a specific pH, as was demonstrated in the exercise. This is crucial in many biochemical processes where specific pH levels are required for optimal system performance.

pH regulation

pH regulation is crucial in both biological and chemical processes as it ensures that environments remain stable for biochemical reactions. For biological systems, such as intracellular fluids, maintaining a specific pH is vital for processes like enzyme activity and metabolic functions to occur properly.

Buffers are the primary means of achieving pH regulation. They consist of a weak acid and its conjugate base, or a weak base and its conjugate acid. The unique feature of buffers is their ability to resist pH changes upon the addition of small amounts of acid or base. This ability is because the buffer components neutralize added acids or bases, thus stabilizing the pH.

For instance, the phosphate buffer is effective in maintaining the pH of intracellular fluids around 7.1 to 7.2 due to its components, \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\), which are present in concentrations that can easily adjust to changes in hydrogen ion concentration, preserving the delicate balance needed for cellular function.

Acid-base equilibrium

Acid-base equilibrium refers to the balance between acids and bases in a solution, which dictates the pH level. It is a dynamic state where the forward and reverse reactions of acid dissociation occur at equal rates, maintaining a constant concentration of the acid and its conjugate base.

The equilibrium is essential because it impacts how a solution can respond to changes, such as the addition of other acids or bases. In a phosphate buffer, the equilibrium between \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\) alongside hydrogen ions is crucial for sustaining the pH level within the narrow range necessary for cellular processes.

When an acid is added to this system, \(\mathrm{HPO}_{4}^{2-}\) ions can react with the excess \(\mathrm{H}^{+}\) ions to form \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\), thus minimizing the increase in \(\mathrm{H}^{+}\) and maintaining pH. Conversely, if additional base is introduced, \(\mathrm{H}_{2}\mathrm{PO}_{4}^{-}\) donates a hydrogen ion to counter the added base, again stabilizing the pH. This ability of the buffer to maintain equilibrium under different conditions is what makes it central to pH regulation in biological systems.