Question

What is the ph of the endpoint of the titration of a 25.00mL sample of .10M HAc (60.05g/mol) using .10M NaOH as the titrant? Ka for HAc is 1.75x10^-5

Answer 1

  The initial pH, before the addition of strong base.

            In dealing with weak acids and bases, there will be an incomplete dissociation in water.  The degree of dissociation will be determined by K_a , the acid dissociation constant in this case.

            CH₃COOH D CH₃COO^- + H⁺

            K_a =     [H⁺] [CH₃COO^-]

                           [CH₃COOH]

            In this example, we have a 0.1 M solution of CH₃COOH, so the [H⁺] can be calculated using the equilibrium expression:

            K_a = 1.8 x 10^-5 =     [x] [x]   , where x is the amount of weak acid that dissociates.

                                           [0.1-x]

            x = 1.34 x 10^-3  = [H⁺]             pH = 2.87

            The pH is higher in this problem than in the strong acid titration because we have partial dissociation of the acid, and

            consequently, a lower [H⁺].

2.         The pH after addition of 10 mL of base.

            We need to determine the moles of CH₃COOH remaining and the volume of the solution to calculate [H⁺].

            moles = [concentration] x volume

            moles OH^- =  (0.1 mol/L) x (0.010 L) = 0.0010 mol OH^-

            starting moles CH₃COOH = (0.1 mol/L) x (0.025 L) = 0.0025 mol CH₃COOH

            moles CH₃COOH remaining after addition of 10 mL base = 0.0025 – 0.0010 = 0.0015 mol CH₃COOH

            volume after addition of 10.0 mL base = 25.0 mL acid + 10.0 mL base = 35 mL = 0.035 L

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

            and produce the conjugate base, CH₃COO^- and H₂O.  At this point in the curve, we have neutralized 0.0010 moles of the acid and produced 0.0010 moles of CH₃COO^-.  These are the new startingconcentrations for the equilibrium process.  A reaction table         (ICE) can be set up to calculate the new equilibrium concentrations.

                                    [CH₃COOH] (M)        [CH₃COO^-] (M)          [H⁺] (M)

            Initial                0.0015mol/35 mL         0.0010 mol/35 mL        0.0 mol/35 mL

            Change             – x                               +x                                +x

            Equilibrium       0.043 – x                     0.029 + x                     x

            K_a = 1.8 x 10^-5 = (0.029 + x) (x)          assume that x is << 0.043 or 0.029       1.8 x 10^-5 = (0.029)(x)             x = 2.67 x 10^-5

                                         (0.043 – x)                                                                                             (0.043)

                                               

            x = [H⁺]           pH = - log (2.67 x 10^-5) = 4.57

           

            An alternative approach is to realize that these are buffering conditions, with the weak acid, CH₃COOH, and its conjugate base,             CH₃COO^-, both present in the solution.  These conditions are appropriate for use of the Henderson-Hasselbalch Equation.

            pH = pK_a + log( [A^-]/[HA])

            At this point in the titration, [A^-] = 0.029 M and [HA] = 0.043 M.  The pK_a = - log (1.6 x 10^-5) = 4.75.

            pH = 4.75 + log (0.029/0.043) = 4.75 -.17 = 4.57

            Comparing this to the pH value of the solution of strong acid after addition of 10 mL of strong base (pH = 1.37) illustrates the     common ion effect.  The presence of one of the products of the dissociation reaction, CH₃COO^-, lessens the extent of      dissociation, and leads to a much higher pH, since fewer H⁺ ions are formed.

3.         The pH after addition of 20 mL of base.

           

            The same steps are followed, as outlined above.

            moles OH^- = 0.0020 = moles CH₃COO^-

            moles CH₃COOH remaining = 0.0025 – 0.0020 = 0.0005

            volume = 25.0 mL + 20.0 mL = 45.0 mL

            pH = pK_a + log ([A^-]/[HA])      pH = 4.75 + log (0.044/0.11) = 5.35

4.         pH at the equivalence point

            moles OH^- = 0.0025 = moles CH₃COO^-

            moles CH₃COOH remaining = 0.0025 – 0.0025 = 0

            At this point, the only species present is CH₃COO^-, the conjugate base of a weak acid.  It is a hydrolyzing species, which will         react with water to make a basic solution.

            CH₃COO^- + H₂O D CH₃COOH + OH^-          K_b = K_w / K_a = 1 x 10^-14/ 1.8 x 10^-5 = 5.56 x 10^-10

                                    [CH₃COO^-]                 [CH₃COOH]               [OH^-]

            Initial                0.0025 mol/ 50 mL       0.0 mol             0.0 mol

            Change             – x                               + x                               + x

            Equilibrium       0.05 – x                        x                                 x

            5.56 x 10^-10 =      x²                              x = 5.27 x 10^-6             pOH = 5.28     pH = 14.00 – 5.28 = 8.72

                                    (0.05 – x)

            At the equivalence point, we will have a basic solution in the titration of a weak acid with a strong base.   In the strong acid/strong         base titration, we had a neutral solution at the equivalence point.

5.         pH after addition of 35.0 mL of base

           

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

            moles OH^- added = 0.0035 moles

            moles CH₃COOH reacting with OH^‑ = 0.0025

            moles OH^- remaining – 0.0035 – 0.0025 = 0.0010

            volume = 25.0 mL + 35.0 mL = 60.0 mL

            [OH^-] = 0.0010 mol / 0.060 L = 0.01667 M

            There will be an equilibrium between the added OH^- and the CH₃COO^-, but the added base will predominate and will determine

            the pH of the solution, as it did in the strong acid/strong base titration.

            pOH = - log (1.67 x 10^-2) = 1.78          pH = 12.22

, [CH₃COOH] = [CH₃COO^-], the pH = pK_a.  In this titration the half way point comes after addition of 12.5 mL of base.  The buffering region extends between ± 1 pH unit from the pK_a.  In this case, from pH = 3.75 – 5.75.