Question

A 200 mL solution of .100 M solution of Ammonia, NH3, is slowly mixed with .100 M HCl. Determine the pH after addition of: 0 ml of HCl, 50 mL of HCl, 100 mL of HCl, 200 mL of HCl, and 250 mL of HCl.

Answer 1

1. The initial pH, before the addition of strong acid.

In dealing with weak acids and bases, there will be an incomplete dissociation in water.

The degree of dissociation will be determined by Kb, the base dissociation constant in this case.

             NH3 + H2O <---> NH4+ + OH-

             Kb = [NH4+][OH-]/[NH3]
In this example, we have a 0.1 M solution of NH3, so the [OH-] can be calculated using the equilibrium expression:

             Kb = 1.8 x 10-5 =[x][x]/[0.1-x],   x is the amount of weak base that dissociates.

            x = 1.34 x 10-3 = [OH-]                     
           pOH = 2.87   
           pH = 14.00 – 2.87 = 11.13
          
2. The pH after addition of 50 mL of 0.1 M HCl.

We need to determine the moles of NH3 remaining and the volume of the solution to calculate [H+].

            moles = [concentration] x volume
            moles H+ = (0.1 mol/L) x (0.050 L) = 0.0050 mol H+
          
           starting moles NH3 = (0.1 mol/L) x (0.200 L) = 0.020 mol NH3
            moles NH3 remaining after addition of 50 mL acid = 0.02 – 0.005 = 0.015 mol NH3

            volume after addition of 50.0 mL acid = 200 mL base + 50.0 mL acid = 250 mL = 0.250 L
          
   Again, we will use the equilibrium expression to calculate the [H+]. The added acid will neutralise some of the bases, and produce the conjugate acid, NH4+ and OH-. At this point in the curve, we have neutralised 0.005 moles of the base and produced 0.005 moles of NH4+. These are the new starting concentrations for the equilibrium process.
           A reaction table (ICE) can be set up to calculate the new equilibrium concentrations.
          
                           [NH3] (M)                    [NH4+] (M)                  [OH-] (M)

            Initial         0.015 mol/0.250L         0.005 mol/0.250L               0.0 mol/0.250L

            Change             – x                               +x                      +x

            Equilibrium       0.06 – x                     0.02 + x                     x
          
          
          
           Kb = 1.8 x 10-5 = (0.02 + x) (x) /(0.074 – x)    assume that x is << 0.06 or 0.02
          
           1.8 x 10-5 = (0.02)(x)/0.06
           x = [OH-] = 5.40 x 10-5
          
           pOH = - log (5.40 x 10-5) = 4.26

           pH = 14.00 – 4.26 = 9.74
          
   3. The pH after addition of 100 mL of acid.

           The same steps are followed, as outlined above.
            moles H+ = 0.010 = moles NH4+
           [NH4+] = 0.010/0.3 L = 0.033
            moles NH3 remaining = 0.02 – 0.01 = 0.01
           [NH3] = 0.01/0.3 = 0.033
             volume = 200 mL + 100 mL = 300 mL = 0.3 L

            pOH = pKb + log ([BH+]/[B])            
           pOH = 4.75 + log (0.033/0.033) = 4.75
           pH = 14- 4.75 = 9.25
   4. pH at the equivalence point

            moles H+ = 0.02 = moles NH4+
            moles NH3 remaining = 0.02 – 0.02 = 0

            At this point, the only species present is NH4+, the conjugate acid of a weak base.
           It is a hydrolyzing species, which will react with water to make an acidic solution.

             NH4+ + H2O <---> NH3 + H3O+
           Ka = Kw / Kb = 1 x 10-14/ 1.8 x 10-5 = 5.56 x 10-10

                                     [NH4+]                         [NH3]                           [H+]

            Initial                0.02 mol/ 400 mL                  0.0 mol                        0.0 mol

            Change                        – x                            + x                               + x

            Equilibrium                  0.05 – x                        x                                 x

            5.56 x 10-10 =      x2 /(0.05 – x)    

           x = 5.27 x 10-6          

           pH = 5.28     

5.pH after addition of 250 mL of HCl

            After the equivalence point, there will be an excess of acid, which will determine the pH.

            moles H+ added = 0.025 moles

            moles NH3 reacting with H+ = 0.02

            moles H+ remaining = 0.025 – 0.020 = 0.005

            volume = 200 mL + 250 mL = 450 mL

            [H+] = 0.005 mol / 0.45 L = 0.0111 M

            There will be an equilibrium between the added H+ and the NH3, but the added acid will predominate.

             pH = - log (0.0111) = 1.955