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 a 1.0 \(M\) 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.)

Short answer

To obtain a pH 6.50 buffer, add 0.984 L of the 1.0 M NaOH solution to 1.0 L of the 1.0 M H2A solution.

Step-by-step solution

Write the dissociation reactions

The dissociation reactions for the given weak diprotic acid H2A are: \(1st \ dissociation: \mathrm{H}_{2}\mathrm{A\longleftrightarrow H^{+} + HA^{-}}\) \(2nd \ dissociation: \mathrm{HA}^{-}\longleftrightarrow \mathrm{H^{+}+A^{2-}}\) Since we are mainly concerned with the first dissociation step, we will focus on the reaction involving \(\mathrm{H}_{2}\mathrm{A}\) and \(\mathrm{HA}^{-}\).

Set up the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is given by: \(pH = pK_a + \log \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) We are given that the buffer should have a pH of 6.50 and the \(K_{a1} = 2 \times 10^{-2}\). Now, we need to find the amount of \(\mathrm{NaOH}\) to be added to form the pH 6.50 buffer. \(p_{K_{a1}} = -\log K_{a1}\) Calculate \(p_{K_{a1}}\): \(p_{K_{a1}} = -\log(2 \times 10^{-2}) \approx 1.70\)

Calculate the required ratio of [A-]/[HA]

Now, plug the pH and \(p_{K_{a1}}\) values into the Henderson-Hasselbalch equation and solve for the ratio \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\): \(6.50 = 1.70 + \log \frac{[\mathrm{A^-}]}{[\mathrm{HA}]}\) Calculate the ratio of [A-]/[HA]: \(\frac{[\mathrm{A^-}]}{[\mathrm{HA}]} = 10^{(6.50-1.70)} \approx 10^{4.8} \approx 63.1\)

Relate the ratio to the amount of NaOH

The reaction between \(\mathrm{H}_{2}\mathrm{A}\) and \(\mathrm{NaOH}\) is: \(\mathrm{H}_{2}\mathrm{A+NaOH\to HA^{-}+H_2O+Na^{+}}\) Let x moles of \(\mathrm{NaOH}\) be added to 1.0 L of 1.0 M \(\mathrm{H}_{2}\mathrm{A}\). According to the reaction, x moles of \(\mathrm{HA}^{-}\) and x moles of \(\mathrm{Na^{+}}\) will be formed. After the reaction, [A-] = x M and [HA] = (1.0 M - x) M. Now, we can use the ratio calculated in step 3 to find x: \(\frac{x}{1-x} = 63.1\)

Calculate the moles of NaOH needed

Solve for x: \(x = \frac{63.1}{63.1 +1} = \frac{63.1}{64.1} \approx 0.984 M\) Since the volume of the acid solution is 1.0 L, the moles of NaOH required would be: Moles of NaOH = (Molarity of NaOH) x (Volume of the solution in liters) Moles of NaOH = 0.984 moles/L x 1.0 L = 0.984 moles

Find the volume of NaOH solution to be added

Recall that the \(\mathrm{NaOH}\) solution is at a concentration of 1.0 M. To find the volume of the \(\mathrm{NaOH}\) solution that needs to be added, divide the moles of \(\mathrm{NaOH}\) by its concentration: \(Volume \ of \ NaOH \ solution = \frac{0.984 \ moles \ NaOH}{1.0 \ M \ NaOH} = 0.984 \ L\) Therefore, to obtain a pH 6.50 buffer, you should add 0.984 L of the 1.0 M NaOH solution to 1.0 L of the 1.0 M H2A solution.

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a key tool in chemistry that relates pH, pKa, and the concentrations of the acid and its conjugate base in a solution. This equation is especially useful for calculating the pH of buffer solutions, which are solutions that resist changes in pH upon the addition of small amounts of acid or base.
The equation is defined as follows: \[ pH = pK_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right) \] Where:

• \(pH\) is the measure of acidity or basicity of the solution.
• \(pK_a\) is the negative logarithm of the acid dissociation constant \(K_a\), reflecting the strength of the weak acid.
• \([\mathrm{A^-}]\) represents the concentration of the conjugate base.
• \([\mathrm{HA}]\) is the concentration of the weak acid.

To use the Henderson-Hasselbalch equation effectively, first calculate the \(pK_a\) from the given \(K_a\) of your acid. Then, plug in the desired pH and the \(pK_a\) to solve for the ratio of \([A^-]\) to \([HA]\). This ratio will guide you on how much base or acid to add to your solution to create the desired buffer.

Acid-base reactions

Acid-base reactions are fundamental chemical reactions involving the transfer of protons (\(H^+\)) between reactants. These reactions occur between an acid, which donates a proton, and a base, which accepts a proton. In the context of buffer solutions, these reactions help maintain a consistent pH by neutralizing added acids or bases.
For the diprotic acid \(\mathrm{H_2A}\) in the given problem, the relevant acid-base reaction involves its first dissociation: \[ \mathrm{H_2A \longleftrightarrow H^+ + HA^-} \] When \(\mathrm{NaOH}\) is added to the acidic solution, the \(\mathrm{OH^-}\) ions from \(\mathrm{NaOH}\) react with \(\mathrm{H^+}\) ions, forming water and decreasing \(\mathrm{H^+}\) concentration: \[ \mathrm{H_2A + NaOH \rightarrow HA^- + H_2O + Na^+} \] The reaction shifts towards the formation of \(\mathrm{HA^-}\), which is crucial in forming the buffer since \(\mathrm{HA^-}\) acts as the conjugate base. By adjusting the amounts of \(\mathrm{HA}\) and \(\mathrm{A^-}\), you modulate the solution's ability to neutralize additional acids or bases.

pH calculation

Calculating pH is essential for understanding the acidity or basicity of a solution. pH is defined as the negative logarithm of the hydrogen ion concentration: \[ pH = -\log[\mathrm{H^+}] \] In buffer systems, the pH calculation goes beyond simply measuring \([\mathrm{H^+}]\). Instead, it involves using the Henderson-Hasselbalch equation to consider both the weak acid and its conjugate base.
For the problem at hand, achieving a buffer with pH 6.50 requires careful adjustment of the concentrations of \([\mathrm{HA}]\) and \([\mathrm{A^-}]\). After determining the \(pK_a\), substitute the known pH and solve for the ratio of these concentrations.
Using the equation: \[ 6.50 = pK_a + \log \left( \frac{[\mathrm{A^-}]}{[\mathrm{HA}]} \right) \] Solve for the ratio, which delivers valuable insight into the amount of base required. The balance of \([\mathrm{HA}]\) and \([\mathrm{A^-}]\) thus dictates the buffer's ability to maintain the desired pH.