Question

Phosphate buffers are important in regulating the pH of intracellular fluids. If the concentration ratio of \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} / \mathrm{HPO}_{4}^{2-}\) in a sample of intracellular fluid is \(1.1 : 1,\) what is the pH of this sample of intracullular fluid? $$ \mathrm{H}_{2} \mathrm{PO}_{4}^{-}(a q) \rightleftharpoons \mathrm{HPO}_{4}^{2-}(a q)+\mathrm{H}^{+}(a q) \quad K_{\mathrm{a}}=6.2 \times 10^{-8} $$

Short answer

Using the Henderson-Hasselbalch equation and the given concentration ratio of \(\mathrm{H_{2}PO_{4}^{-}}\) to \(\mathrm{HPO_{4}^{2-}}\) (1.1 : 1) and the \(K_\mathrm{a}\) value (\(6.2 \times 10^{-8}\)), the pH of the sample of intracellular fluid can be calculated as follows: $$ \mathrm{pH} = 7.21 + \log \frac{1.1}{1} \approx 7.34 $$ Hence, the pH of the sample of intracellular fluid is approximately 7.34.

Step-by-step solution

Identify the given values and write down the Henderson-Hasselbalch equation.

We are given the concentration ratio of \(\mathrm{H_{2}PO_{4}^{-}}\) to \(\mathrm{HPO}_{4}^{2-}\) and the \(K_\mathrm{a}\) value for the reaction. We will use the Henderson-Hasselbalch equation: $$ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]} $$ where pH is the desired value, \(K_\mathrm{a}\) is the acid dissociation constant, [A⁻] is the concentration of the conjugate base (in this case, \(\mathrm{HPO_{4}^{2-}}\)), and [HA] is the concentration of the weak acid (in this case, \(\mathrm{H_{2}PO_{4}^{-}}\)).

Rewrite the concentration ratio as a fraction.

The given ratio is 1.1 : 1, which means there is 1.1 mole of \(\mathrm{H_{2}PO_{4}^{-}}\) for every 1 mole of \(\mathrm{HPO_{4}^{2-}}\). To use in the Henderson-Hasselbalch equation, rewrite it as a fraction: $$ \frac{[\mathrm{H_{2}PO_{4}^{-}}]}{[\mathrm{HPO_{4}^{2-}}]} = \frac{1.1}{1} $$

Convert the given Ka value to pKa.

Take the negative logarithm of the given Ka value (\(6.2 \times 10^{-8}\)) in order to find the pKa: $$ \mathrm{p}K_\mathrm{a} = -\log (6.2 \times 10^{-8}) \approx 7.21 $$

Plug values into the Henderson-Hasselbalch equation.

Now, we have all the values needed to use the Henderson-Hasselbalch equation and solve for the pH: $$ \mathrm{pH} =7.21+ \log \frac{1.1}{1} $$

Calculate the pH.

Perform the calculations in the equation: $$ \mathrm{pH} = 7.21 + \log (1.1) \approx 7.34 $$ The pH of the sample of intracellular fluid is approximately 7.34.

Key concepts

pH Calculation

To determine the pH of a solution, especially in biochemical contexts, the Henderson-Hasselbalch equation is an invaluable tool. It allows us to calculate the pH by using the concentrations of an acid and its conjugate base and the acid's dissociation constant, Ka.
The equation is written as:

• \(\mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\)

Here, \([\mathrm{A}^-]\) represents the concentration of the conjugate base, and \([\mathrm{HA}]\) is the concentration of the acid.
To effectively use this formula, we need the ratio of the concentrations, as given in a problem, and the pKa value derived from the Ka. This equation is crucial because it connects the concept of equilibrium in chemical reactions to measurable properties like pH.

Acid Dissociation Constant

The acid dissociation constant, \(K_a\), is a fundamental value in understanding the strength of an acid in a solution. It quantifies the equilibrium concentration of the dissociated form (products) and the undissociated form (reactants) of an acid reaction.
For example, in the dissociation of \(\mathrm{H}_2\mathrm{PO}_4^-\), we have:

• \(\mathrm{H}_2\mathrm{PO}_4^-(aq) \rightleftharpoons \mathrm{HPO}_4^{2-}(aq) + \mathrm{H}^+(aq)\)

The \(K_a\) expresses how far the reaction proceeds to the right (dissociation) at equilibrium. A small \(K_a\) indicates a weak acid that doesn't dissociate much, implying stronger conjugate base formation.
To find the pKa used in the Henderson-Hasselbalch equation, we take the negative logarithm of the Ka: \(\mathrm{p}K_a = -\log K_a\). This transformation simplifies comparisons and calculations, making the concept of acid strength more user-friendly.

Phosphate Buffer

Phosphate buffers, due to their capacity to resist pH changes, play an essential role in biological systems. They are typically made up of pairs like \(\mathrm{H}_2\mathrm{PO}_4^-\) and \(\mathrm{HPO}_4^{2-}\), which can either donate or accept protons, helping stabilize pH levels in solutions with changes in acidity.
As seen in the provided exercise, the ratio of \(\mathrm{H}_2\mathrm{PO}_4^-/\mathrm{HPO}_4^{2-}\) is crucial. This equilibrium helps in keeping intracellular fluids at a steady state, which is vital for various cellular functions.
Phosphate buffers range effectively around a pH close to their pKa, around 7.21 here, as seen in this equation. This is optimal for processes requiring a neutral pH, making the phosphate system a widespread buffer in biological and chemical applications.

Intracellular Fluid

Intracellular fluid refers to the fluid contained within the cells, making up about 40% of an organism's total body weight. The pH of this fluid is tightly regulated, typically around 7.4, to maintain proper cellular function.
Cells conduct countless reactions, many of which are pH-sensitive. Thus, intracellular fluids rely on buffers, like the phosphate buffer, to prevent fluctuations that could disrupt cellular activities. This buffering system manages to keep cells from becoming too acidic or too basic, which could lead to enzyme denaturation or metabolic disturbances.
Understanding how to calculate the pH of intracellular fluid using concepts like buffering and the Henderson-Hasselbalch equation helps elucidate how living systems maintain homeostasis and respond to environmental changes efficiently.