Question

Suppose you want to do a physiological experiment that calls for a pH \(6.5\) buffer. You find that the organism with which you are working is not sensitive to the weak acid \(\mathrm{H}_{2} \mathrm{X}\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) or its sodium salts. You have available a \(1.0 \mathrm{M}\) solution of this acid and a \(1.0 \mathrm{M}\) solution of \(\mathrm{NaOH}\). How much of the \(\mathrm{NaOH}\) solution should be added to \(1.0 \mathrm{~L}\) of the acid to give a buffer at \(\mathrm{pH}\) 6.50? (lgnore any volume change.)

Short answer

In order to create a pH 6.5 buffer using the weak acid H2X and NaOH, you need to add approximately 0.613 L of the 1.0 M NaOH solution to the 1.0 L of 1.0 M H2X solution. This is calculated using the Henderson-Hasselbalch equation and taking into account the second dissociation of the weak acid.

Step-by-step solution

Identify the buffer system

In this exercise, you are using the weak acid H2X with its sodium salts and NaOH to create the buffer. The first dissociation of H2X is represented by: \[ H2X \rightleftharpoons H^+ + HX^- \] The second dissociation is represented by: \[ HX^- \rightleftharpoons H^+ + X^{2-} \]

Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is as follows: \[ pH = pK_a + log \frac{[A^-]}{[HA]} \]

Determine which dissociation to utilize given the desired pH

Compare the pH value of 6.5 to the pKa values calculated for both dissociations: \( pK_{a1} = -\log (2 \times 10^{-2}) = 1.70 \) \( pK_{a2} = -\log (5 \times 10^{-7}) = 6.30 \) As the desired pH is closer to pK_{a2}, we will use the second dissociation of the weak acid.

Use the Henderson-Hasselbalch equation to find the ratio

Now, plug the desired pH and the appropriate pKa value into the Henderson-Hasselbalch equation: \( 6.5 = 6.30 + log \frac{[X^{2-}]}{[HX^-]} \) After solving for the ratio, we find: \[ \frac{[X^{2-}]}{[HX^-]} = 1.585 \]

Determine the amount of NaOH needed

Since initially you have 1.0 L of H2X at 1.0 M concentration, when 1 mole of HX^- reacts with NaOH, 1 mole of X^{2-} will be produced. Let x be the number of moles of NaOH needed, then we can write: \[ \frac{[X^{2-}]}{[HX^-]} = \frac{x}{(1-x)} \] Now, solve for x: \[ x = 1.585 \times (1-x) \] \[ x = \frac{1.585}{2.585} \approx 0.613 \: moles \]

Calculate the volume of NaOH needed

Since the concentration of NaOH is 1.0 M, you can now find the volume of NaOH solution needed using the moles of NaOH: \[ volume\: of\: NaOH = \frac{moles\: of\: NaOH}{concentration\: of\: NaOH} \] \[ volume\: of\: NaOH = \frac{0.613\: moles}{1.0\: M} \] \[ volume\: of\: NaOH \approx 0.613\: L \] Thus, approximately 0.613 L of the 1.0 M NaOH solution should be added to the 1.0 L of 1.0 M H2X solution to achieve a pH of 6.50, accounting for the second dissociation.

Key concepts

Understanding the Henderson-Hasselbalch Equation

In the realm of buffer solutions, it's crucial to familiarize oneself with the Henderson-Hasselbalch equation. This formula is at the heart of understanding how a buffer maintains a stable pH. It relates the pH of the solution to the pKa (the logarithmic measure of the acid dissociation constant) and the ratio between the concentration of the deprotonated species, or the base (\text{[A^-]}), and the concentration of the non-deprotonated species, or the acid (\text{[HA]}).

\textbf{The equation is given by:} \[ pH = pK_a + \text{log} \frac{[A^-]}{[HA]} \]
When preparing a buffer solution, you aim for a particular pH. To reach this goal, you can alter the concentrations of \text{[A^-]} and \text{[HA]} until the equation balances out to give you the desired pH. An understanding of logarithms is essential here, as changes in the ratio of concentrations can correspond to significant shifts in pH. The closer the desired pH is to the pKa of the weak acid, the more effective the buffer will be at resisting changes in pH.

The Process of pH Calculation

Calculating pH is a common task in chemistry that involves determining the acidity or basicity of a solution. A pH less than 7 indicates an acidic solution, while a pH greater than 7 signals a basic solution. The pH can be calculated by taking the negative logarithm to base 10 of the hydrogen ion concentration (\text{[H+]}) in the solution: \[ pH = -\text{log} [H^+] \]
However, when working with a buffer solution, using the Henderson-Hasselbalch equation becomes more practical. The equation takes advantage of the relationship between the pH and the amounts of acid and conjugate base present in the solution. In the context of the problem at hand, you are tasked with adjusting the amounts of weak acid and conjugate base to arrive at a specific pH, which in this case is 6.5. A thorough grasp of pH calculations is key to effectively managing buffer systems in laboratory settings.

Weak Acid Dissociation: A Balancing Act

Weak acid dissociation is a balance between the undissociated acid and its dissociated ions in a solution. Weak acids do not fully dissociate in water, which is what makes them integral to buffer systems—their ability to donate hydrogen ions to the solution is limited and controlled. The degree of this dissociation is quantified by the acid dissociation constant (\text{K}_\text{a}), and it's an intrinsic property of the acid.

In solutions of weak acids, the dissociation can be represented by an equilibrium expression. For instance, for an acid \text{HA} dissociating into \text{H^+} and \text{A^-}: \[ HA \rightleftharpoons H^+ + A^- \]
The acid dissociation constant is then expressed by: \[ K_a = \frac{[H^+][A^-]}{[HA]} \]
When preparing a buffer, you make use of this dissociation. By choosing an acid with a suitable \text{K}_\text{a}, and by adding its conjugate base or a strong base such as NaOH, you adjust the pH to the desired level. The weak acid dissociation, coupled with the Henderson-Hasselbalch equation, provides the necessary framework to create a buffer with a precise pH—a fundamental process in various biological and chemical experiments.