For ammonia, K_b is 1.8 *10^-⁵ . To make a buffered solution with pH 10.0, the ratio of NH₄Cl to NH₃ must be

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The Henderson-Hasselbalch equation for a buffered solution is: pH = pKa + log([A-]/[HA]) We are given that the pKa for ammonia is 9.25. To create a buffered solution with a pH of 10.0, we need to choose the ratio of [A-] to [HA] so that: 10.0 = 9.25 + log([A-]/[HA]) 0.75 = log([A-]/[HA]) Antilog(0.75) = [A-]/[HA] 5.62 = [A-]/[HA] We are also given that K_b is equal to: K_b = [NH4+][OH-]/[NH3] Since we want to create a buffered solution using NH₄Cl and NH₃, we can rewrite K_b in terms of these two species: K_b = ([NH3] - [NH3]x)/([NH3]x + [NH4+]x) * ([OH-]x + [Cl-]) / ([OH-]x) where [NH3]x and [NH4+]x are the concentrations of ammonia and ammonium ions in the buffered solution, and [OH-]x and [Cl-]x are the concentrations of hydroxide and chloride ions in the buffered solution. Since the pH of the buffered solution is 10.0, we know that the concentration of [OH-]x is 10^(-4), and we can calculate the concentration of [NH3]x using the Henderson-Hasselbalch equation: 5.62 = [A-]/[HA] = [NH3]x/[NH4+]x [NH3]x = 5.62[NH4+]x pH = pKa + log([A-]/[HA]) 10.0 = 9.25 + log([NH3]x/[NH4+]x) 0.75 = log(5.62) [NH3]x/[NH4+]x = 5.62 Substituting these values into the expression for K_b, we get: K_b = ([NH3]x - [NH3]x^2/[NH4+]x) * ([OH-]x + [Cl-]) / ([OH-]x) 1.8 * 10^(-5) = ([NH3]x - [NH3]x^2/[NH4+]x) * (10^(-4) + [Cl-]x) / (10^(-4)) Solving for [Cl-]x/[NH4+]x, we get: [Cl-]x/[NH4+]x = 0.18 Therefore, the ratio of NH₄Cl to NH₃ must be 0.18 : 1.

For Ammonia, Kb Is 1

For Ammonia, K_b Is 1