Calculate the \(\mathrm{pH}\) of a solution that is \(0.40 \mathrm{M} \mathrm{H}_{2} \mathrm{NNH}_{2}\) and \(0.80 M \mathrm{H}_{2} \mathrm{NNH}_{3} \mathrm{NO}_{3} .\) In order for this buffer to have \(\mathrm{pH}=\) \(\mathrm{p} K_{\mathrm{a}},\) would you add HCl or NaOH? What quantity (moles) of which reagent would you add to \(1.0 \mathrm{L}\) of the original buffer so that the resulting solution has \(\mathrm{pH}=\mathrm{p} K_{\mathrm{a}} ?\)

First, we'll need to find the Ka (acid dissociation constant) value for the amine cation, \(\mathrm{H}_{2}\mathrm{NNH}_{3}^+\). Recall that Ka is related to Kb by the equation Ka = Kw/Kb, where Kw is the ionization constant of water (\(1.0 \times 10^{-14}\)). In this case, we are given that \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) is a \(0.40 M\) weak base, and, as an amine, it has a Kb value of \(4.0 \times 10^{-7}\). To find Ka, we'll use the relationship mentioned above: Ka = Kw/Kb = \(\frac{1.0 \times 10^{-14}}{4.0 \times 10^{-7}}\) Now, we calculate the Ka and pKa values: \[Ka = \frac{1.0 \times 10^{-14}}{4.0 \times 10^{-7}} = 2.5 \times 10^{-8}\] Taking the negative logarithm of Ka gives us the pKa value: \[pKa = -\log(2.5 \times 10^{-8}) \approx 7.60\]

Now that we have the pKa value, we can apply the Henderson-Hasselbalch equation to calculate the pH of the buffer solution: \[pH = pKa + \log \left(\frac{[\mathrm{A}^-]}{[\mathrm{HA}]}\right)\] Here, \(\mathrm{A^-}\) represents the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) (the weak base), and \(\mathrm{HA}\) represents the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{3}^+\) (the weak acid) which can be approximated to be equal to the concentration of \(\mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\). The given concentrations are as follows: \([A^-] = 0.40 M \) \([HA] = 0.80 M \) Substituting the values into the equation: \[pH = 7.60 + \log \left(\frac{0.40}{0.80}\right)\] Calculating the pH: \[pH = 7.60 + \log(0.5) \approx 7.10\]

Since the pH of our buffer is 7.10, and to make the pH equal to the pKa (7.60), we need to increase the pH. To do this, we would need to add NaOH (a strong base). Now let's find the moles of NaOH required to adjust the pH. Starting with the Henderson-Hasselbalch equation and rearranging for \(\frac{[A^-]}{[HA]}\): \[\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 10^{(pH-pKa)}\] We know the pH (\(7.60\)) and the pKa (\(7.60\)) values, so we can solve for the ratio: \[\frac{[\mathrm{A}^-]}{[\mathrm{HA}]} = 10^{(7.60-7.60)} = 10^0 = 1\] This means the concentrations of \(\mathrm{H}_{2}\mathrm{NNH}_{2}\) and \(\mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\) should be equal, and since the solution volume is \(1.0 L\), the moles of each species should also be equal. Since we initially have \(0.40 \, mol \mathrm{H}_{2}\mathrm{NNH}_{2}\) and \(0.80 \, mol \mathrm{H}_{2}\mathrm{NNH}_{3}\mathrm{NO}_{3}\), we need to add \(0.80 - 0.40 = 0.40 \, mol\) of NaOH to the solution to achieve a 1:1 ratio. In summary, the pH of the given buffer solution is 7.10. To change the pH to the pKa value of 7.60, we need to add 0.40 moles of NaOH to 1.0 L of the buffer solution.