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The Henderson-Hasselbalch equation is given by: \[ \mathrm{pH} = \mathrm{p}K_\mathrm{a} + \log \frac{[\mathrm{A}^-]}{[\mathrm{HX}]} \] where pH is the pH of the solution, pKa is the acid dissociation constant of the acidic species (NH4+), [A-] is the concentration of the conjugate base (NH3), and [HX] is the concentration of the conjugate acid (NH4+).

To calculate the ratio \(\frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^+\right]}\) for each pH value, we will rearrange the Henderson-Hasselbalch equation to isolate the ratio and then substitute the given pH values and the known pKa value (9.25) for the NH3/NH4+ system. \[ \frac{[\mathrm{NH}_{3}]}{[\mathrm{NH}_{4}^+]} = 10^{\mathrm{pH} - \mathrm{p}K_\mathrm{a}} \] a. For pH = 9.00: \[ \frac{[\mathrm{NH}_{3}]}{[\mathrm{NH}_{4}^+]} = 10^{9.00 - 9.25} = 10^{-0.25} \approx 0.56 \] b. For pH = 8.80: \[ \frac{[\mathrm{NH}_{3}]}{[\mathrm{NH}_{4}^+]} = 10^{8.80 - 9.25} = 10^{-0.45} \approx 0.36 \] c. For pH = 10.00: \[ \frac{[\mathrm{NH}_{3}]}{[\mathrm{NH}_{4}^+]} = 10^{10.00 - 9.25} = 10^{0.75} \approx 5.62 \] d. For pH = 9.60: \[ \frac{[\mathrm{NH}_{3}]}{[\mathrm{NH}_{4}^+]} = 10^{9.60 - 9.25} = 10^{0.35} \approx 2.24 \] So, the ratio of the concentrations of ammonia to ammonium ion for each given pH value is approximately: a. pH = 9.00: \(\frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^+\right]} \approx 0.56\) b. pH = 8.80: \(\frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^+\right]} \approx 0.36\) c. pH = 10.00: \(\frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^+\right]} \approx 5.62\) d. pH = 9.60: \(\frac{\left[\mathrm{NH}_{3}\right]}{\left[\mathrm{NH}_{4}^+\right]} \approx 2.24\)