Question

If 0.175 L of 0.364 M ammonia solution is titriated with 0.095 L of 0.35 M HCI, what is the pH of the resulting solution?

Answer 1

The problem involves a weak base (NH3) and strong acid (HCl). When NH3 is mixed with HCl we are left with an equilibrium as follows:

NH3 + HCl NH4⁺ + Cl^-

NH3 is a weak base so its conjugate base NH4⁺ will be strong. Similarly HCl is a strong acid so it will a weak conjugate acid as Cl^-.

In order to find out the pH we need to first write the number of moles of each of the above species mentioned.

(number of moles = concentration * volume)

               NH3        +    HCl            NH4⁺    +                    Cl^-

(initial) n(NH3) = 0.364 * 0.175                n(HCl) = 0.35 * 0.095                  -                                                -

              = 0.06                                             = 0.03

Now HCl is acting as the limiting reagent in above reaction and will get completely consumed in the reaction. So now the final amount left after the reaction is as follows

(final) n(NH3) = 0.06 - 0.03 = 0.03        n(HCl) = 0.03 - 0.03 = 0            n(NH4⁺) = 0.03                      n(Cl^-) = 0.03

Now from all above as HCl is completely consumed therefore it is not going to affect the pH of the solution. Cl- is also a weak base of HCl so it plays no role to affect the pH. Only left over NH3 and NH4+ plays role. So the equation involving both is as follows by using which, we will find out the concenration of OH- in order to further find pH. Here is the dissociation constant of NH3, Kb = 1.8 * 10^-5 plays an important role.

        NH3                 + H2O                    NH4⁺ +                  OH^-

         C(NH3) = n/ v_total                 -                                    C(NH4⁺) = 0.03/0.27= 0.11

v_total = total volumeof the solution = 0.27

(initial) C(NH3) = 0.03/0.27 = 0.11

(final) C(NH3) = 0.11 - x                                                  C(NH4⁺) = 0.11 + x                                       +x

Using dissociation constant of base the equilibrium equation can be written as follows:

Kb = {[NH4] * [OH^-]} / [NH3]

     substituting the above concentrations

Kb = {(0.11 + x) * (x)} / (o.11 - x)

    1.8 * 10^-5 = 0.11x/0.11

So x = [OH-] = 1.8*10^-5

Hence p[OH] = 4.74

pH = 14 - p[OH] = 14 - 4.74 = 9.26