What stoichiometric concentration of the indicated substance is required to obtain an aqueous solution with the pH value shown: (a) aniline, \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{2}\), for \(\mathrm{pH}=8.95 ;(\mathrm{b}) \mathrm{NH}_{4} \mathrm{Cl}\) for \(\mathrm{pH}=5.12 ?\)

The Henderson-Hasselbalch equation is given by: \[pH = pKa + log \left(\frac{[A^-]}{[HA]}\right)\] where \(pH\) is the -log of the H+ concentration, \(pKa\) is the -log of the dissociation constant (Ka) of the acid, [A^-] is the concentration of the base (in moles per liter), and [HA] is the concentration of the acid (in moles per liter).

This equation applies when the weak species gaining the proton (aniline in this case) is a weak base. The equation becomes: \[pOH = pKb + log \left(\frac{[BH+]}{[B]}\right)\] Where \(B\) is the weak base (aniline in this case), and \(BH+\) is the protonated form of the base. As we are given the pH and not the pOH, we will need to calculate the pOH first using the equation \( pOH = 14 - pH\). Given that \(pH=8.95\), the pOH becomes \(14 - 8.95 = 5.05\). Aniline's \(pKb = 9.37\). Therefore, rearranging the equation to solve for the ratio \(\left(\frac{[BH+]}{[B]}\right)\) gives: \(5.05 = 9.37 + log\left(\frac{[BH+]}{[B]}\right)\) Solving for [B] (unprotonated aniline) yields: \(\frac{[BH+]}{[B]}=10^{(5.05 - 9.37)}=6.22x10^{-5}\) Identifying the stoichiometric concentration would involve setting equal amounts of \(BH+\) and \(B\) and then setting the ratio equal to the one found above.However, as [B]=[BH+], we can simply take the inverse of the ratio found before to receive the initial stoichiometric concentration which amounts to \(\frac{1}{6.22x10^{-5}}=1.61×10^{-4} M\)

Now, let's apply the Henderson-Hasselbalch equation to \(NH_4Cl\). The active species here is \(NH_4^+\), which acts as a weak acid. Therefore, the equation appears as originally defined: \(pH = pKa + log(\frac{[A^-]}{[HA]})\). The \(pKa\) for \(NH_4^+\) is \(9.25\). Given that \(pH=5.12\), this can be rearranged to find the [A-]/[HA] ratio: \(5.12 = 9.25 + log(\frac{[A^-]}{[HA]})\) Therefore, \([A^-]/[HA] = 10^{(5.12 - 9.25)} = 1.46 × 10^{-4}\). As \(NH_4Cl\) dissociates fully and all \(Cl^-\) ions are spectators, the \(NH_4^+\) ions will be of the same concentration as \(Cl^-\) ions, hence we can argue that [A-]=[HA] after dissociation. Setting these equal, the stoichiometric concentration can be found by simply taking the inverse of the ratio found before which gives \(\frac{1}{1.46 × 10^{-4}} = 6.85 × 10^3 M\) for the initial stoichiometric concentration of \(NH_4Cl\).