Question

Draw titration curve if titrate 10.00 mL of 0.250 M CH3NH2 with 0.125 M HCl. Also, calculaute pH@ 0.0 mL, 5.00 mL, 10.00 mL, 20.00 mL, 20.50 mL

Please explain everything you are doing so I could understand how you solved. Do not write in cursive.

Answer 1

Again, we will use the equilibrium expression to calculate the [H⁺].  The added acid will neutralize some of the base,

and produce the conjugate acid, CH₃NH₃⁺ and OH^-.  At this point in the curve, we have neutralized 0.000625 moles of the base and produced 0.000625 moles of CH₃NH₃⁺.  These are the new starting concentrations for the equilibrium process.  A reaction table can be set up to calculate the new equilibrium concentrations.

                                    [CH₃NH₂] (M)                    [CH₃NH₃⁺] (M)                  [OH^-] (M)

            Initial                0.001875mol/15 mL         0.000625 mol/15 mL        0.0 mol/15 mL

            Change             – x                               +x                                +x

            Equilibrium       0.125 – x                     0.042 + x                     x

            K_b = 4.4 x 10^-4 = (0.042 + x) (x) /(0.125-x) assume that x is << 0.125 or 0.042

x = 7.42 x 10^-3  

  

            x = [OH^-]         pOH = - log (7.42 x 10^-3) = 2.13 pH = 14.00 – 2.13 = 11.87

           

            An alternative approach is to realize that these are buffering conditions, with the weak base, CH₃NH₂, and its conjugate acid, CH₃NH₃⁺, both present in the solution.  These conditions are appropriate for use of the Henderson-Hasselbalch Equation.

            pOH = pK_b + log( [BH⁺]/[B])

            At this point in the titration, [BH⁺] = 0.042 M and [B] = 0.125 M.  The pK_b = - log (4.4 x 10^-4) = 3.36

            pOH = 3.36 + log (0.042/0.125) = 2.13 pH = 11.87