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 \(\mathrm{H}_{2} \mathrm{X}\) \(\left(K_{a 1}=2 \times 10^{-2} ; K_{a 2}=5.0 \times 10^{-7}\right)\) 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 \(\mathrm{pH} 6.50 ?\) (Ignore any volume change.)

We first need to calculate the pKa of the weak acid involved. To do this, we use the relationship between \(K_{a}\) and pKa: pKa = \(-\log(K_{a})\). From the problem, we are given \(K_{a1} = 2 \times 10^{-2}\). Using this value: \[ \mathrm{pKa}_{1} = -\log(2 \times 10^{-2}) \] \[ \mathrm{pKa}_{1} \approx 1.70 \]

Next, we use the Henderson-Hasselbalch equation to relate the pH, pKa, and the ratio of the acidic and basic forms in the buffer system: \[ \mathrm{pH} = \mathrm{pKa} + \log_{10} \left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right) \] With \(\mathrm{pH} = 6.50\) and \(\mathrm{pKa} \approx 1.70\), we insert these values into the equation: \begin{equation} 6.50 = 1.70 + \log_{10} \left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right) \end{equation}

Now, we will solve the equation (1) from step 2 to find the ratio of the conjugate base \(\mathrm{X}^{-}\) to the acid \(\mathrm{HX}\): \begin{equation} \log_{10} \left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right) = 6.50 - 1.70 \end{equation} \[ \log_{10} \left(\frac{[\mathrm{Base}]}{[\mathrm{Acid}]}\right) = 4.8 \] To find the ratio, we will take the antilog of the value calculated: \[ \frac{[\mathrm{Base}]}{[\mathrm{Acid}]} = 10^{4.8} \] \[ \frac{[\mathrm{X}^{-}]}{[\mathrm{HX}]} \approx 63000 \]

Since the buffer solution has a volume of \(1.0\mathrm{~L}\), we can assume the original acid concentration is \(1.0\mathrm{~M}\). After reaction with NaOH, we can define the final concentrations of the acid and conjugate base as follows: \[ [\mathrm{HX}] = 1.0 - x \ \mathrm{M} \] \[ [\mathrm{X}^{-}] = x \ \mathrm{M} \] Now using the ratio found in step 3: \[ \frac{x}{1.0 - x} = 63000 \] Solve for x: \[ x \approx 0.9999\ \mathrm{M} \] Thus, the amount of \(\mathrm{NaOH}\) to be added is: \[ \mathrm{Volume\ of\ NaOH} = \frac{0.9999\ \mathrm{moles}}{1.0\ \mathrm{M}} \] \[ \mathrm{Volume\ of\ NaOH} \approx 1.0\ \mathrm{L} \] Therefore, you would need to add roughly \(1.0\ \mathrm{L}\) of \(\mathrm{NaOH}\) solution to the acid to create a buffer solution with a pH of 6.50. Note: Since we ignored the volume change, the result is an approximation.