Problem 95 - Chemistry The Central Science In Si Units

Question

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

Step-by-Step Solution

Answer

Add 0.613 liters of \( \mathrm{NaOH} \) to the solution.

Step 1: Identify the Relevant Equilibrium

We are creating a buffer with a pH of 6.50 using a diprotic acid \( \mathrm{H}_{2}\mathrm{A} \), characterized by \( K_{a1} = 2 \times 10^{-2} \) and \( K_{a2} = 5 \times 10^{-7} \). Given that \( pH = 6.50 \), we should consider the dissociation involving \( K_{a2} \) because it is closer to \( pH = 6.50 \). We use the equilibrium \( \mathrm{HA}^{-} \longleftrightarrow \mathrm{H}^{+} + \mathrm{A}^{2-} \).

Step 2: Apply Henderson-Hasselbalch Equation

For the reaction involving \( K_{a2} \), the Henderson-Hasselbalch equation is:\[ pH = pK_{a2} + \log \left( \frac{[\mathrm{A}^{2-}]}{[\mathrm{HA}^{-}]} \right) \]Calculate \( pK_{a2} \):\[ pK_{a2} = -\log(K_{a2}) = -\log(5 \times 10^{-7}) = 6.30 \]Then, substitute \( pH = 6.50 \):\[ 6.50 = 6.30 + \log \left( \frac{[\mathrm{A}^{2-}]}{[\mathrm{HA}^{-}]} \right) \]

Step 3: Calculate the Ratio of Base to Acid Forms

Rearrange to find the ratio:\[ \log \left( \frac{[\mathrm{A}^{2-}]}{[\mathrm{HA}^{-}]} \right) = 6.50 - 6.30 = 0.20 \]Convert the log expression:\[ \frac{[\mathrm{A}^{2-}]}{[\mathrm{HA}^{-}]} = 10^{0.20} \approx 1.58 \]

Step 4: Determine Amount of \(\mathrm{NaOH}\) Required

Initially, we have 1.0 L of 1.0 M \( \mathrm{H}_{2}\mathrm{A} \), so we have 1.0 mol of \( \mathrm{H}_{2}\mathrm{A} \). Upon reaction with \( \mathrm{NaOH} \), it donates protons to form \( \mathrm{HA}^- \) and \( \mathrm{A}^{2-} \). Apply the ratio:\[ [\mathrm{HA}^{-}] = \frac{1}{1 + 1.58}(1 \text{ mol}) \approx 0.387 \text{ mol} \]\[ [\mathrm{A}^{2-}] = 1.0 \text{ mol} - 0.387 \text{ mol} = 0.613 \text{ mol} \]

Step 5: Calculate the Volume of \(\mathrm{NaOH}\)

\( \mathrm{NaOH} \) will neutralize one proton per mole. Therefore, the amount of \( \mathrm{NaOH} \) required to convert \( \mathrm{HA}^{-} \) to \( \mathrm{A}^{2-} \) is 0.613 mol.Since we have a 1 M \( \mathrm{NaOH} \) solution, the volume of \( \mathrm{NaOH} \) needed is:\[ V = \frac{0.613}{1.0} = 0.613 \text{ L} \]

Key Concepts

Henderson-Hasselbalch Equation

When dealing with buffer solutions, the Henderson-Hasselbalch equation is your best friend. This formula helps relate the pH of a solution to the pKa of the acid and the ratio of concentrations of its deprotonated (base) and protonated (acid) forms. The equation is:

\[pH = pK_a + \log \left( \frac{[A^-]}{[HA]} \right)\]

In this context:

• \([A^-]\) represents the concentration of the conjugate base.
• \([HA]\) represents the concentration of the acid.
• Using this, you can calculate the pH of a solution if you know the concentrations of the acid and its conjugate base.

It's especially handy when designing buffer solutions to maintain a desired pH. In our exercise, this equation allows us to find out how much NaOH to add to the diprotic acid to achieve a specific pH of 6.50.

Diprotic Acid

Diprotic acids are exciting because they can donate two protons (hydrogen ions) per molecule, unlike monoprotic acids which donate just one. In our exercise, the diprotic acid in question is \(\mathrm{H}_2\mathrm{A}\). It dissociates in steps:

• First, it loses one proton to form \(\mathrm{HA}^-\).
• Then, it can lose a second proton to form \(\mathrm{A}^{2-}\).

Each dissociation step has its equilibrium constant, \(K_{a1}\) and \(K_{a2}\). Here, \(K_{a1} = 2 \times 10^{-2}\) and \(K_{a2} = 5 \times 10^{-7}\).
In buffer calculations like ours, you often focus on a specific step of dissociation. The pH and equilibrium constants help you determine which equilibrium is more relevant, deciding which step's dissociation affects the pH the most.

pH Calculation

Calculating pH is crucial for understanding and manipulating solutions in chemistry. pH indicates how acidic or basic a solution is, with its scale ranging from 0 to 14. A value less than 7 means acidic, while a value above 7 indicates a basic solution.

To calculate pH:

• Find the concentration of hydrogen ions.
• Use the formula: \( pH = -\log [H^+] \)

In the buffer solution exercise, the calculated pH helps you determine how much of a base (NaOH in this case) is required to reach a certain pH level. This ties back to using equilibrium constants, showing how balancing acid and base forms allows you to maintain a desired pH in practical applications.

Acid-Base Equilibrium

Acid-base equilibrium is foundational in chemistry, representing the balance between acids and bases in a solution. Equilibrium constants \(K_a\) describe how strongly an acid dissociates and releases hydrogen ions.

Understanding acid-base equilibrium is essential for various reasons:

• It's crucial for predicting the behavior of acids and bases in solution.
• It helps in designing buffer systems, like the one in our exercise, to stabilize pH levels within biological or chemical environments.

In an equilibrium state, the rate of forward reaction (where acid dissociates to release \(H^+\)) matches the rate of the reverse reaction (where the conjugate base recombines with \(H^+\)), maintaining constant concentrations. This provides a stable environment where pH can be maintained, which is especially critical in physiological experiments and other delicate processes.