A 25.00 mL aliquot of 0.100 M NH₃ is titrated with 21.00 mL of 0.0900 M HCl.Calculate the pH.K_b = 1.76 x 10⁻⁵ A)5)25 B)8)44 C)8)75 D)5)56 E)7)00

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First, we can calculate the number of moles of HCl used in the titration. n(HCl) = c(HCl) * V(HCl) = 0.0900 M * 21.00 mL = 0.00189 mol Since NH₃ is a weak acid, it undergoes partial dissociation, so we need to use the equilibrium expression to find the concentration of H+ ions. NH₃ + H2O ⇄ H3O+ + N₃− At equilibrium, let the concentration of NHSA be x. We can assume that the initial concentration of NHSA is equal to the concentration after dissociation, x. We also know that the equilibrium concentration of N₃− will be equal to the initial concentration of NHSA. Therefore, [N₃−] = x. Using the equilibrium expression for Ka, we can set up an equation to solve for x. Ka = [H3O+][N₃−] / [NHSA] 1.76 x 10⁻⁵ = x^2 / (0.100 - x) Since we are dealing with a weak acid, we can assume that x is much smaller than 0.100, so we can simplify the expression to: 1.76 x 10⁻⁵ = x^2 / 0.100 Solving for x, we get: x = 0.0103 M Now we can use the definition of pH to find the pH of the solution. pH = -log[H+] [H+] = x = 0.0103 M pH = -log(0.0103) = 1.02 Rounding to 2 decimal places, the pH of the solution is 1.02. The closest answer choice to this value is C, 8.75.

A 2500 ML Aliquot of 0