The pH of a 0.25 M aqueous solution ammonia,NH₃,at 25.0 °C is 9.50.What is the value of K_b for NH₃?

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The equation for the dissociation of ammonia in water is: NH3 (aq) + H2O (l) ⇌ NH4+ (aq) + OH- (aq) The equilibrium constant expression for this reaction is: K_b = [NH4+][OH-]/[NH3] Since the pH of the solution is 9.50, we know that the concentration of hydroxide ions is: [OH-] = 10^-pH = 10^-9.50 = 3.16 × 10^-10 Assuming that the concentration of NH4+ is negligible compared to the concentration of NH3, we can write: K_b ≈ [OH-][NH4+]/[NH3] Substituting the known values into the expression, we get: K_b ≈ (3.16 × 10^-10)(x)/(0.25 - x) where x is the concentration of NH4+ in mol/L. Since NH3 is a weak base, we can assume that the concentration of NH3 doesn't change significantly upon dissociation, meaning that 0.25 - x ≈ 0.25. Using the pKa of NH4+ (9.24), we can solve for x: Kb = Kw/Ka = 1.0 × 10^-14/1.8 × 10^-5 = 5.6 × 10^-10 x^2/(0.25) = Kb x = sqrt(Kb * 0.25) = 4.7 × 10^-3 Substituting x into the expression for K_b, we get: K_b ≈ (3.16 × 10^-10)(4.7 × 10^-3)/(0.25) K_b ≈ 5.9 × 10^-9 The closest answer choice is (C), 4.0 × 10^-9.

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