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 \(\mathrm{M}\) solution of \(\mathrm{NaOH} .\) How much of the NaOH solution should be added to 1.0 \(\mathrm{L}\) of the acid to give a buffer at pH 6.50\(?\) (Ignore any volume change.)

The Henderson-Hasselbalch equation is given by: pH = pKa + log10(\(\dfrac{[A^{-}]}{[HA]}\)) where pH is the desired pH of the buffer, pKa is the logarithm of the acid dissociation constant (Ka), [A^-] is the concentration of the base (conjugate of the weak acid), and [HA] is the concentration of the weak acid.

We are given two dissociation constants for the weak acid \(\mathrm{H}_{2} \mathrm{A}\). We need to choose the one that will be suitable for preparing the buffer with the desired pH. The choice should be made based on the difference between the pH and pKa values. The pKa value closest to the desired pH should be chosen. The first dissociation constant is \(K_{a_1} = 2 \times 10^{-2}\). We can find the pKa as follows: pKa1 = -log10(\(K_{a_1}\)) = -log10(2 * 10^-2) ≈ 1.7 The second dissociation constant is \(K_{a_2} = 5.0 \times 10^{-7}\). We can find the pKa as follows: pKa2 = -log10(\(K_{a_2}\)) = -log10(5 * 10^-7) ≈ 6.3 Since pKa2 (6.3) is closer to the desired pH (6.50) than pKa1 (1.7), we choose pKa2.

Now that we have chosen the appropriate pKa value, we can use the Henderson-Hasselbalch equation to find the amount of NaOH needed: 6.50 = 6.3 + log10(\(\dfrac{[A^{-}]}{[HA]}\)) 0.2 = log10(\(\dfrac{[A^{-}]}{[HA]}\)) Now, we must find the ratio of the base to the acid: 10^0.2 = \(\dfrac{[A^{-}]}{[HA]}\) 1.585 = \(\dfrac{[A^{-}]}{[HA]}\) Since initially, we have 1 L of 1.0 M weak acid solution, the initial moles of [HA] can be calculated as follows: Initial moles of [HA] = 1.0 L * 1.0 M = 1.0 mol Let x be the moles of NaOH added to the acid solution. When x moles of NaOH are added, x moles of [HA] will be converted into [A^-]: [HA] = 1.0 - x [A^-] = x Now we can plug these values into our ratio equation: 1.585 = \(\dfrac{x}{1.0 - x}\) Now solve for x: x = 1.585 * (1.0 - x) x = 1.585 - 1.585 * x 2.585 * x = 1.585 x ≈ 0.613 mol

Since the concentration of NaOH solution is 1.0 M, we can use the formula: Moles of NaOH = Concentration of NaOH * Volume of NaOH 0.613 mol = 1.0 M * Volume of NaOH Volume of NaOH ≈ 0.613 L Therefore, to create a buffer with the pH of 6.50, we need to add approximately 0.613 L of 1.0 M NaOH solution to 1.0 L of the 1.0 M weak acid solution.