Question

Calculate the hydronium ion concentration and the pH when 50.0ml of 0.40M NH3 is mixed with 50.0ml of 0.40M Hcl

Answer 1

Ans. Number of moles of NH3 = (Molarity x Vol in L ) of NH3 solution

                                                = 0.40 M x 0.050 L = 0.02 moles

Number of moles of HCl = (Molarity x Vol in L ) of HCl solution

                                                = 0.40 M x 0.050 L = 0.02 moles

NH3(aq) + HCl(aq) -----------> NH4Cl(Aq)       - balanced reaction.

According to the stoichiometry, 1 mol NH3 completely reacts with 1 mol HCl. Since, number of moles of NH3 is exactly equal to that of HCl, both solutions neutralize each other with NH4Cl(aq) as product.

At equilibrium, the solution consists of only NH4Cl (aq).

So, number of moles of NH₄⁺ = 0.02

            Total volume of solution = 100.0 mL

Concentration of NH₄⁺ = moles of NH₄⁺ / total volume of solution

                                    = 0.02 moles / 0.100 L = 0.2 M

However, NH₄⁺, a cation of the salt of weak base and strong acid, donates protons to the aqueous medium (see reaction below). The resultant solution, therefore becomes, acidic because of H⁺ ions (= H₃O⁺, hydronium ion) present into the solution.

                        NH₄⁺(aq) + H2O(l) ----------> H₃O⁺(aq) + NH3(aq)     

Initial                 0.2                                            0                      0

Change             -x                                             +x                    +x

Equilibrium        (0.2-x)                                       x                       x

Now, dissociation constant, Ka of NH₄⁺               [use, Ka(NH4+) = 5.6 x 10^-10]

            Ka = ([NH3] [H3O+] ) / (NH4+)

            Or, 5.6 x 10^-10 = x² / (0.2 - x)

            Or, x² = 0.2 x 5.6 x 10^-10                          [Assuming, (0.2 – x) = 0.2 because x << 0.2 ]

            Or, x = 1.05 x 10^-5

Thus, at equilibrium, [H₃O⁺] = x = 1.05 x 10^-5

pH = - log (1.05 x 10^-5) = 4.97