Question

A 10.0−mL solution of 0.330 M NH₃ is titrated with a 0.110 M HCl solution. Calculate the pH after the following additions of the HCl solution:

(a) 0.00 mL



(b) 10.0 mL



(c) 30.0 mL



(d) 40.0 mL

Answer 1

Given that The volume of NH_3 = 10.0 ml The molarity of NH_3 = 0.3 M The number of moles of NH_3 = Molarity * volume in liters = 0.3 M * 0.01 L = 0.003 mol The molarity of HCl = 0.10 M We know that Kb of NH_3 = 1.8 * 10^- 5 pKb = 4.74 The stoichiometric equation for the reaction between HCl and NH_3 can be written as HCl(aq) + NH_3(aq) rightarrow NH_4Cl(aq) + H_2O(l) This equation shows one mole of HCl converts one mole of NH_3 into one mole of NH_4 Cl a) The volume of HCl = 0.0 ml Therefore in this case the solution contains only ammonia We know that [OH^-] = squareroot Kb times C Where C = concentration of ammonia = 0.30 M [OH^-] = squareroot Kb times C = 1.8 times 10^- 5 times 0.3 M = 2.3238 * 10^- 3 M