Question

The \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}-\mathrm{HPO}_{4}^{2-}\) combination plays a role in maintaining the pH of blood. (a) Write equations to show how a solution containing these ions functions as a buffer. (b) Verify that this buffer is most effective at \(\mathrm{pH} 7.2\) (c) Calculate the \(\mathrm{pH}\) of a buffer solution in which \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}\right]=0.050 \mathrm{M}\) and \(\left[\mathrm{HPO}_{4}^{2-}\right]=0.150 \mathrm{M} .[\)Hint: Focus on the second step of the phosphoric acid ionization.]

Short answer

a) \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-} \leftrightarrow \mathrm{H^{+}} + \mathrm{HPO}_{4}^{2-}\). b) The buffer is indeed most effective at \(\mathrm{pH} 7.2\). c) The \(\mathrm{pH}\) comes out to approximately 7.5 after calculation.

Step-by-step solution

Part (a): Buffer equation formulation

Since this is a buffer solution, there would be an equilibrium between the \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\) and \(\mathrm{HPO}_{4}^{2-}\). The reactions can be written as follows: \[ \mathrm{H}_{2} \mathrm{PO}_{4}^{-} \leftrightarrow \mathrm{H^{+}} + \mathrm{HPO}_{4}^{2-}\]

Part (b): Verification of effectiveness at pH 7.2

To verify the effectiveness of the buffer at \(\mathrm{pH} 7.2\), one needs to use the Henderson-Hasselbalch equation, \[ \mathrm{pH} = \mathrm{pKa} + \log \left(\frac{\left[\mathrm{Base}\right]}{\left[\mathrm{Acid}\right]}\right)\]. Generally, a buffer is most effective when \(\mathrm{pH} = \mathrm{pKa}\), which implies that the molar concentrations of the Acid and the conjugate Base are equal. In this case, pKa2 for \(\mathrm{H}_{2} \mathrm{PO}_{4}^{-}/\mathrm{HPO}_{4}^{2-}\) is approximately 7.2, hence a buffer using these two species would indeed be most effective at a pH of 7.2.

Part (c): pH calculation

The \(\mathrm{pH}\) is calculated again using the Henderson-Hasselbalch equation from part (b). Given that \(\left[\mathrm{H}_{2} \mathrm{PO}_{4}\right]=0.050 \mathrm{M}\) and \(\left[\mathrm{HPO}_{4}^{2-}\right]=0.150 \mathrm{M}\), one can substitute these values: \[ \mathrm{pH} = 7.2 + \log \left(\frac{\left[\mathrm{HPO}_{4}^{2-}\right]}{\left[\mathrm{H}_{2} \mathrm{PO}_{4}^{-}\right]}\right)=7.2+\log \left(\frac{0.15 \mathrm{M}}{0.05 \mathrm{M}}\right) \].

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is key to solving problems involving buffer solution pH calculation. This equation provides an easy way to estimate the pH of a buffer solution based on the concentrations of an acid and its conjugate base.

It is given typically as: \[\begin{equation}\t\text{pH} = \text{pKa} + \text{log} \bigg(\frac{[\text{Base}]}{[\text{Acid}]}\bigg)\end{equation}\]In this context, pKa represents the acid dissociation constant, a measure of the strength of the acid. The ratio of the concentrations of the base and acid essentially shows how much the acid has ionized. The closer the pH of the buffer is to the pKa value of the acid, the more effective the buffer is at resisting changes in pH. This is why a buffer is most effective when the pH is equal to the pKa, and the concentrations of the acid and base are equal.

When applying this equation to calculate pH, like in the solution for part (c), you first need to know the pKa value and the molar concentrations of the acid and base in the buffer. The logarithmic term in the equation accounts for the ratio between the conjugate base and the acid which determines the pH value. The simplicity of the Henderson-Hasselbalch equation lies in its power to use simple arithmetic to address a complex chemical balance.

Acid-Base Equilibrium

Acid-base equilibrium is a critical concept in understanding how buffer systems work. It denotes the balance between an acid and its conjugate base in an aqueous solution. When an acid donates a proton (H+), it forms its conjugate base; conversely, when a base accepts a proton, it forms its conjugate acid.

The reaction can reach a dynamic equilibrium where the rate of the forward reaction, the acid losing a proton, is equal to the rate of the reverse reaction, the base gaining a proton. The equilibrium position is represented by the Ka (acid dissociation constant) value, which quantifies the strength of an acid in solution. A strong acid has a high Ka value, indicating it dissociates more in water, releasing more H+ ions, whereas a weak acid has a lower Ka value. Similarly, pKa is the negative logarithm of Ka and offers another way to express the strength of an acid.

In buffer solutions, a weak acid and its conjugate base (or a weak base and its conjugate acid) are present in ratio that allows for the stabilization of the pH upon addition of small amounts of other acids or bases. The concept of acid-base equilibrium helps us understand how a buffer maintains the pH within a narrow range, an essential quality for many biochemical processes.

Phosphoric Acid Ionization

Phosphoric acid (H3PO4) is a triprotic acid, meaning it can donate three protons, with each ionization step having its own pKa value. During ionization, phosphoric acid can form dihydrogen phosphate (\[\begin{equation}\tH_2PO_4^{-}\end{equation}\]), hydrogen phosphate (\[\begin{equation}\tHPO_4^{2-}\end{equation}\]), and finally phosphate (\[\begin{equation}\tPO_4^{3-}\end{equation}\]) ions in a stepwise manner as protons are progressively lost. Understanding these ionization steps and their respective equilibrium constants is crucial for predicting the behavior of phosphate buffers in solution.

For the exercise given, the second ionization step is of interest: \[\begin{equation}\tH_2PO_4^{-} \leftrightarrow H^+ + HPO_4^{2-}\end{equation}\], where \[\begin{equation}\tH_2PO_4^{-}\end{equation}\] is acting as an acid, and \[\begin{equation}\tHPO_4^{2-}\end{equation}\] is the conjugate base. The effectiveness of the phosphate buffer can be described by the pKa2 value of the H2PO4-/HPO42- pair, which is around 7.2, close to the physiological pH, making it an important buffer system in biological contexts.

Buffer System in Blood

The buffer system in blood is a vital mechanism that helps to maintain the pH within the narrow range necessary for physiological function. Among the various buffer systems in the body, the bicarbonate buffer system is the most prominent in the blood, where it regulates the pH by an equilibrium between carbonic acid (H2CO3) and bicarbonate ions (HCO3-). However, the phosphate buffer system, consisting of \[\begin{equation}\tH_2PO_4^{-}\end{equation}\] and \[\begin{equation}\tHPO_4^{2-}\end{equation}\], also plays a role, although to a lesser extent.

The presence of \[\begin{equation}\tH_2PO_4^{-}\end{equation}\] and \[\begin{equation}\tHPO_4^{2-}\end{equation}\] can absorb excess H+ ions or donate H+ ions to neutralize added bases, thereby resisting changes in pH. In this way, the phosphate buffer systems, along with other buffer systems, work to keep the blood pH around 7.4, which is ideal for the function of hormones, enzymes, and other cellular activities. The capacity to maintain this stable environment is crucial, as even slight variations in blood pH can have profound effects on overall health and biological processes.