17.3 A Closer Look at Acid-Base Titrations

A titration is an experiment in which the concentration of
    an unknown acid or base is deduced from the
    concentration of a standard solution of base or acid.

Consider the addition of a standard NaOH solution added to
    a solution of HCl.
A pH meter is used to measure the acidity of the solution.

As a 0.100 M NaOH solution is added to a 0.100 M solution
    of HCl, the pH increases from the starting point of 1.00.

As more base is added, the pH gradually increases.
Near the equivalence point (the point at which the acid and
    base are present in stoichiometric quantities), the pH
    begins to change dramatically with small additions of
    base.

The last few drops of base before the equivalence point
    change the pH drastically from around 3.00 to 7.00.

At the equivalence point, the pH is EXACTLY 7.00, because
    the solution is neutral, (all the acid has been neutralized
    by the added base).  All that is present at this point in
    solution is the salt, NaCl, and H₂O.  This is called the
    equivalence point, and NOT the endpoint.

THEREFORE:  A SALT DERIVED FROM STRONG ACID AND
    STRONG BASE, IN WATER, HAS A pH = 7.00.
Examples:  KNO₃, RbI, BaBr₂, LiCl, SrClO₄

Addition of one drop of NaOH after equivalence causes the
    pH to increase dramatically to about 10.00.

If we add the indicator, phenolphthalein, it changes from
    colorless to light red right after the equivalence point,
    at a pH of about 8.3.

The end point of the titration is the point at which we
    observe a color change, which is at pH about 8.3 with
    phenolphthalein as the indicator.

The end point is experimentally determined, the
    equivalence point is determined by stoichiometry.
 

Strong Acid-Strong Base Titrations
The plot of pH versus volume during a titration is a
    titration curve.  Consider adding a solution of a strong
    base (e.g., 0.1 M NaOH) to a solution of a strong acid
    (e.g., 0.10 M HCl).

WE ARE GOING TO CALCULATE THE pH OF THE SOLUTION
AT VARIOUS POINTS ALONG THE TITRATION:

1)   100.0 mL OF 0.100 M HCl (no NaOH SOLUTION
        ADDED!!)          pH = -log(0.100 M) = 1.00

2)   ADD 10.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING  THE HCl.
       # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                      = 10.00 mmoles
       # MOLES OF NaOH ADDED = 10.00 mL X 0.100 M
                                                    = 1.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 9.00
       mmoles OF HCl LEFT IN A TOTAL VOLUME OF 110 mL.

    pH = -log(9.00 mmoles/110 mL) = -log(.08182) = 1.09

3)   ADD 28.50 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HCl.
      # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
      # MOLES OF NaOH ADDED = 28.50 mL X 0.100 M
                                                   = 2.85 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 7.15
      mmoles OF HCl LEFT IN A TOTAL VOLUME OF 128.5 mL.

  pH = -log(7.15 mmoles/128.5 mL) = -log(.05564) = 1.25

4)   ADD 90.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HCl.
       # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                      = 10.00 mmoles
       # MOLES OF NaOH ADDED = 90.00 mL X 0.100 M
                                                    = 9.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE IS 1.00
        mmole OF HCl LEFT IN A TOTAL VOLUME OF 190 mL.

 pH = -log(1.00 mmoles/190 mL) = -log(.005263) = 2.28
 

5)   ADD 99.90 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HCl.
       # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                      = 10.00 mmoles
       # MOLES OF NaOH ADDED = 99.90 mL X 0.100 M
                                                    = 9.99 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.01
     mmoles OF HCl LEFT IN A TOTAL VOLUME OF 199.9 mL.

  pH = -log(0.01 mmoles/199.9 mL) = -log(.00005003)
        = 4.30

6)   ADD 100.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HCl.
       # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                      = 10.00 mmoles
       # MOLES OF NaOH ADDED = 100.00 mL X 0.100 M
                                                    = 10.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
        mmoles OF HCl AND 0.00 mmoles OF NaOH LEFT IN
        A TOTAL VOLUME OF 200 mL.
 
        THIS IS THE EQUIVALENCE POINT!!!
        pH = 7.00

7)   ADD 110.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HCl.
       # MOLES OF HCl INITIALLY = 100.0 mL X 0.100 M
                                                      = 10.00 mmoles
       # MOLES OF NaOH ADDED = 110.00 mL X 0.100 M
                                                    = 11.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
        mmole OF HCl LEFT AND 1.00 mmoles OF NaOH IN
        EXCESS IN A TOTAL VOLUME OF 210 mL.

Before any base is added, the pH is given by the strong acid
solution.   Therefore, pH < 7.00.

When base is added, before the equivalence point, the pH is
given by the amount of strong acid in excess.
Therefore, pH < 7.00.

At equivalence point, the amount of base added is
    stoichiometrically equivalent to the amount of acid
    originally present.  Therefore, the pH is determined by
    the salt solution.  Therefore, pH = 7.00.

To detect the equivalence point, we use an indicator that
    changes color somewhere near 7.00.
Usually, we use phenolphthalein, which changes color
    at pH 8.3 and just above the equivalence point at
    pH = 7.00.

In acid, phenolphthalein is colorless.
As NaOH is added, there is a slight pink color at the
    addition point.
When the flask is swirled and the reagents mixed, the
    pink color disappears.
At the end point, the solution remains light pink.

If more base is added, the solution turns a darker pink.

The equivalence point in a titration is the point at which the
    acid and base are present in stoichiometric quantities.

The end point in a titration is the observed color change.

The difference between the equivalence point and the end
    point is called the titration error.

If we add a strong acid to a strong base, the shape of the
    titration curve is very similar to that of a strong base
    added to a strong acid, but the mirror image.

Initially, the strong base is in excess, so the pH > 7.00.
As acid is added, the pH decreases but is still greater than
    7.00.
At equivalence point, the pH is given by the salt solution
    (i.e., pH = 7.00).
After equivalence point, the pH is given by the strong acid
    in excess, so pH < 7.00.
 - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Weak Acid-Strong Base Titrations

Consider the titration of acetic acid, CH₃COOH and NaOH.
Before any base is added, the solution contains only weak
acid.  Therefore, the pH is given by the equilibrium
calculation for just the weak acid.
   CH₃COOH (aq) + H₂O(l) ----> H₃O⁺(aq) + CH₃COO^-(aq)
                                            <----

               K_a  = [H₃O⁺][CH₃COO^-]/[CH₃COOH]
 

As strong base is added, the strong base consumes a
    stoichiometric quantity of weak acid:

  CH₃COOH (aq) + NaOH(aq) -----> CH₃COO^-(aq) + H₂O(l)

However, at this point, there is a mixture of weak acid
    and its conjugate base = BUFFER.
The pH is given by the buffer calculation.

First, the amount of CH₃COO^- generated is calculated, as
well as the amount of CH₃COOH consumed. (Stoichiometry)

Then the pH is calculated using equilibrium conditions
        (or use Henderson-Hasselbalch equation.)
 
 

WE ARE GOING TO CALCULATE THE pH OF A SOLUTION
    AT VARIOUS POINTS ALONG THE TITRATION:

1)   100.0 mL OF 0.100 M CH₃COOH(HA)   (NO NaOH
        SOLUTION ADDED!!)

        K_a  = [H₃O⁺][A^-]/[HA] = 1.8 x 10^-5 = x²/0.100-x
           x = 0.00134 M = [H₃O⁺]

         pH = -log(0.00134 M) = 2.87
 

2)   ADD 10.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING  THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLES OF NaOH ADDED = 10.00 mL X 0.100 M
                                                    = 1.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 9.00
    mmoles OF HA AND 1.00 mmoles OF NaA PRODUCED
    IN A TOTAL VOLUME OF 110 mL.
        THEREFORE, THIS IS A BUFFER!!

       K_a  = [H₃O⁺][A^-]/[HA] = 1.8 x 10^-5
             = (x)(1.00mmoles/110mL)/(9.00mmoles/110mL)

          x = 0.000162 M = [H₃O⁺]

        pH = -log(0.000162) = 3.79

3)   ADD 28.50 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLES OF NaOH ADDED = 28.50 mL X 0.100 M
                                                    = 2.85 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 7.15
    mmoles OF HA AND 2.85 mmoles OF NaA PRODUCED IN
    A TOTAL VOLUME OF 128.5 mL.
            THEREFORE, BUFFER!!

    K_a = [H₃O⁺][A^-]/[HA] = 1.8 x 10^-5
       = (x)(2.85mmoles/128.5mL)/(7.15mmoles/128.5mL)

      x = 0.0000452 M = [H₃O⁺]

    pH = -log(0.0000452) = 4.34
 

4)   ADD 90.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLES OF NaOH ADDED = 90.00 mL X 0.100 M
                                                    = 9.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 1.00
    mmole OF HA AND 9.00 mmoles OF NaA PRODUCED IN
    A TOTAL VOLUME OF 190 mL.
            THEREFORE, BUFFER!!

      K_a  = [H₃O⁺][A^-]/[HA] = 1.8 x 10^-5
            = (x)(9.00/190)/(1.00/190)

         x = 0.000002 M = [H₃O⁺]

       pH = -log(0.000002) = 5.70
 

5)   ADD 99.90 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLES OF NaOH ADDED = 99.90 mL X 0.100 M
                                                    = 9.99 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.01
    mmoles OF HA AND 9.99 mmoles OF NaA PRODUCED IN
    A TOTAL VOLUME OF 199.9 mL.
            THEREFORE, BUFFER!!

     K_a = [H₃O⁺][A^-]/[HA] = 1.8 x 10^-5
          = (x)(9.99/199.9)/(0.01/199.9)

       x = 0.000000018 M = [H₃O⁺]

    pH = -log(0.000000018) = 7.74

NOTE!!  THE pH IS ABOVE 7.00 AND WE ARE NOT YET AT
    THE EQUIVALENCE POINT!!
 

6)   ADD 100.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLES OF NaOH ADDED = 100.00 mL X 0.100 M
                                                    = 10.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
    mmoles OF HA AND 10.00 mmoles OF NaA PRODUCED IN
    A TOTAL VOLUME OF 200 mL.
              THIS IS THE EQUIVALENCE POINT!!!

           pH IS NOT = 7.00

HYDROLYSIS OF SALT OF WEAK ACID!!

    [SALT] = 10.00 mmoles/200 mL = 0.050 M

   [OH^-] = (K_w/K_a)([SALT])]^1/2

   [OH^-] = [(1 x 10^-14/1.8 x 10^-5)(0.050 M)]^1/2
              = 0.00000527 M

      pOH = -log(0.00000527) = 5.28
 

   Therefore, pH = 14.00 - 5.28 = 8.72, well above 7.00!!
 

7)   ADD 110.00 mL OF 0.100 M NaOH TO BEAKER
       CONTAINING THE HA.
       # MOLES OF HA INITIALLY = 100.0 mL X 0.100 M
                                                     = 10.00 mmoles
       # MOLE OF NaOH ADDED = 110.00 mL X 0.100 M
                                                  = 11.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
    mmole OF HA LEFT AND 1.00 mmoles OF NaOH IN
    EXCESS IN A TOTAL VOLUME OF 210 mL.

     pOH = -log(1.00 mmoles OH^-/210 mL)
             = -log(.004762) = 2.32

        pH = 14.00 - 2.32 = 11.68

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

At the equivalence point, all the acetic acid has been
    consumed and all the NaOH has been consumed.
    However, CH₃COO^- has been generated.
Therefore, the pH is given by the [CH₃COO^-] solution.
This means pH > 7.00

More importantly, for a weak acid-strong base titration
  pH at the equivalence point is NOT 7.00, but is > 7.00!!!

Let us titrate the weak base  NH₃ by the strong acid HCl

WE ARE GOING TO CALCULATE THE pH OF A SOLUTION
    AT VARIOUS POINTS ALONG THE TITRATION:

1)   100.0 mL OF 0.100 M NH₃ (BOH)   (NO HCl
        SOLUTION ADDED!!)

             NH_{3  +} H₂O  _---->  NH₄⁺+ OH^-
                                       <----

        K_b  = [NH₄⁺][OH^-]/[ NH₃] = 1.8 x 10^-5 = x²/0.100-x
           x = 0.00134 M = [OH^-]

         pOH = -log(0.00134 M) = 2.87;
         therefore, pH = 14.00 - 2.87 = 11.13
 

2)   ADD 40.00 mL OF 0.100 M HCl TO BEAKER
       CONTAINING  THE weak base.
       # MOLES OF base INITIALLY = 100.0 mL X 0.100 M
                                                        = 10.00 mmoles
       # MOLES OF HCl ADDED = 40.00 mL X 0.100 M
                                                = 4.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 6.00
    mmoles OF weak base left AND 4.00 mmoles OF NH₄⁺
    salt PRODUCED IN A TOTAL VOLUME OF 140 mL.
        THEREFORE, THIS IS A BUFFER!!

         K_b  = [NH₄⁺][OH^-]/[ NH₃] = 1.8 x 10^-5
               = (4.00mmoles/110mL)(x)/(6.00mmoles/110mL)

          x = 0.000027 M = [OH^-]

3)   ADD 100.00 mL OF 0.100 M HCl TO BEAKER
       CONTAINING THE weak base.
       # MOLES OF weak base INITIALLY = 100.0 mL X 0.100 M
                                                                  = 10.00 mmoles
       # MOLES OF HCl ADDED = 100.00 mL X 0.100 M
                                                = 10.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
    mmoles OF weak base left AND 10.00 mmoles OF NH₄⁺
    salt PRODUCED IN A TOTAL VOLUME OF 200 mL.
              THIS IS THE EQUIVALENCE POINT!!!

           pH IS NOT = 7.00

HYDROLYSIS OF SALT OF WEAK BASE!!

    [SALT] = 10.00 mmoles/200 mL = 0.050 M

   [H₃O⁺] = (K_w/K_b)([SALT])]^1/2

   [H₃O⁺] = [(1 x 10^-14/1.8 x 10^-5)(0.050 M)]^1/2
                = 0.00000527 M

4)   ADD 110.00 mL OF 0.100 M HCl TO BEAKER
       CONTAINING THE weak base.
       # MOLES OF weak base INITIALLY = 100.0 mL X 0.100 M
                                                                  = 10.00 mmoles
       # MOLE OF HCl ADDED = 110.00 mL X 0.100 M
                                              = 11.00 mmoles

    AFTER REACTION (INSTANTANEOUS), THERE ARE 0.00
    mmole OF weak base LEFT AND 1.00 mmoles OF HCl IN
    EXCESS IN A TOTAL VOLUME OF 210 mL.

---------------------------------------------------------------
Titration Curves for Weak Acids or Weak Bases:
For a strong acid-strong base titration, the pH begins at
less than 7 and gradually increases as base is added.
Near the equivalence point, the pH increases dramatically.

For a weak acid-strong base titration, the initial pH rise is
    more steep than the strong acid-strong base case.
However, then there is a leveling off due to buffer effects.
The inflection point is not as steep for a weak acid-strong
    base titration.
The shape of the two curves after equivalence point is the
    same because pH is determined by the strong base in
    excess.
Two features of titration curves are affected by the
    strength of the acid:
1. the amount of the initial rise in pH, and
2. the length of the inflection point at equivalence.
    The weaker the acid (smaller Ka), the smaller the
    equivalence point inflection.
    For very weak acids (K_a < 10^-9), it is impossible to
    detect the equivalence point in water solution.
------------------------------------------------------------------

Titrations of Polyprotic Acids
 In polyprotic acids or polyprotic bases, each ionizable
    proton dissociates in steps.
 Therefore, in a titration there are n equivalence points
    corresponding to each ionizable proton.

In the titration of H₃PO₄ with KOH, there are three
theoretical equivalence points:
  1-  the first is the formation of H₂PO₄^-1
          H₃PO₄ + NaOH ----->  NaH₂PO₄ + H₂O
  2-  the second is the formation of HPO₄^2-
          NaH₂PO₄ + NaOH ----->  Na₂HPO₄ + H₂O
  3-  the third is the formaiton of PO₄^3-
          Na₂HPO₄ + NaOH ----->  Na₃PO₄ + H₂O

In the titration of Na₂CO₃ with HCl, there are two
    equivalence points:
1. one for the formation of HCO₃^-1
2. one for the formation of H₂CO₃.

In acidic medium, the equilibrium shifts to left, and
the solution has the color of HIn.  In basic medium,
the equilibrium lies to the right, the In^- predominates
and the solution has the color associated with In^-

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