Titration is the controlled neutralization of a
acid and a base. If we consider the titration of a
**weak acid** by a **strong base**, we can analyze the situation completely
and determine all the concentrations of the aqueous species at any
volume addition of the titrant.

Notice, from the figure below, that the titration of a weak acid with a strong base is different from that of a strong acid/base titration, primarily in the begiinig of the titration, i.e., before the equivalence point. Moreover, the equivalence point itself is shifted upward in pH.

We can

The entries in the tables are linked to the calculation.

25.0 mL of 0.100M HA (Ka=1 x 10^{-5}), titrated with with 0.100M NaOH:

Vol. NaOH |
Total Volume |
% Diss |
n_{[H+]e} |
n_{[OH-]e} |
pH |

0 | 25 | (a) | (a) | (b) | (a) |

10 | 35 | (c) | (c) | (b) | (c) |

20 | 45 | (c) | (c) | (b) | (c) |

22 | 47 | (c) | (c) | (b) | (c) |

24 | 49 | (c) | (c) | (b) | (c) |

25 | 50 | (d) | (e) | (e) | (e) |

26 | 51 | (d) | (g) | (f) | (g) |

28 | 53 | (d) | (g) | (f) | (g) |

30 | 55 | (d) | (g) | (f) | (g) |

40 | 65 | (d) | (g) | (f) | (g) |

50 | 75 | (d) | (g) | (f) | (g) |

Hints for the On-line quiz...

**What, in general, is the pH at the equivalence point for this type of titration?****What about at 1/2 equivalents base added?**

**Calculation (a):**

Before any base is added to the acid, the extent of the dissociation
of the acid determines the H^{+}_{(aq)} concentation through the reaction:

Recognizing that the number o moles of H^{+}_{(aq)} is equal to the
number of moles of A^{-}_{(aq)}, and each is equal to the number of
moles of HA that dissociates gives the following expression:

The number of moles of HA present before titration
is just the initial volume times the
initial concentration or 0.025L x 0.100 M= 2.50 mmol of HA.
Solving for the n_{H}+ by successsive approximation or the quadratic formula
will yield a result of 2.488 x 10^{-2}
millimoles of H^{+}_{(aq)}.

Since the volume is 25.0 mL,
the concentration of H^{+}_{(aq)} of 9.950 x 10^{-4} molar (which is approximately 1.0 x 10
^{-3})

The pH is easily calculated from the H^{+}_{(aq)} concentration through
its' definition:

In this case the
**concentrated weak acid** approximation is adequate
to predict the pH, namely because the % dissociation is so low. The result
of this approximation is:

pH =
(1/2){-log(K_{a}{n_{HA}^{0}/V})} = 3.000

**Calculation (b):**

Once the pH is known, the pOH is easily calculated at 25^{o}C by the relation

pOH = 14 - pH

[OH

n

**Calculation (c)**

During the beginning of the titration, the number of moles of HA in solution will be reduced
**stoichiometrically** by reaction
with a strong base through the reaction:

Therefore,
the addition of strong base to the weak acid
creates a **buffer** just like mixing
the salt of the weak acid to the acid. Is it clear how the equilibrium constant for this neutralization is related
to the pKa of the acid HA (10^{14-pKa}).
After this step takes place the equilibrium that must be considered is
the same one as is step (a)

but there are significant quantities of Both HA and A

- The moles of A
^{-}, n_{(A}-_{)}^{0}, is taken to be the moles of strong base added. - The moles of HA, n
_{HA}^{0}, is taken to be the number of moles of HA acid initially present minus the moles of strong base added. - The volume of the solution is equal to the initial volume of the HA acid plus the volume of the base solution that is added.

As a numerical example, consider the point at which 10.0 mL of base is added.

The equilibrium expression above has but one unknown, so we may solve for the number of moles of H

n_{(H}+_{)} = 5.25 x 10^{-4}, pH = 4.82

Can you think of the approximations that are made at this point? Well we can make a useful one by observing that since this is a buffer, we can use the Henderson-hasselbalch equation for the pH of a Buffer. (What approximations are implicit in the use of 'The Buffer equation'? (n

pH = pKa + log_{10}([A-]/[HA])

The amount the acid has dissosciated is

% Diss = n

**Calculation (d)**

At equivalence, the number of moles of base added is equal to the number
of moles of HA acid initially present,
so all of the acid is converted
to its conjugate base anion. Therefore the dissociation is complete

At equivalence, the solution is an aqueous salt. In fact, it is identical to a 0.050M solution of the salt NaA

A^{-}_{(aq)} + H_{2}O = OH^{-}_{(aq)}
+ HA_{(aq)}

The equilibrium constant of this reaction
is related to the pKa of the acid HA
and the solution of the equilibrium expression for this
reaction will yield the hydroxide concentration, [OH^{-}_{(aq)}].

We may use the **concentrated weak base** approximation,
in exact analogy to the concentrated weak acid approximation,
to determine the [OH^{-}_{(aq)}], whcih is tantemount to
ignoring y in the denominator of the equilibrium expression.
Therefore:

y =
[OH^{-}_{(aq)}] =
(K_{b}
[A^{-}])^{1/2} = 7.07 x 10^{-6} M

pOH
= -log_{10}([OH^{-}_{(aq)}]) =
5.15

pH = 14 - pOH = 8.85

n_{OH-(aq)} =
[OH^{-}_{(aq)}] * V_{total}

n_{H+(aq)} =
[H^{+}_{(aq)}] * V_{total}

**Calculation (f)**

Now we consider only the concentration of OH^{-} coming from the excess
strong acid

**Calculation (g)**
Of course, as the titration has changed into a simple strong base
dilution problem, you have considered and calculated
the pOH first
and use it to determine the pH and n_{(H}+_{)}.

n_{(H}+_{)} = 10^{-pH} * V_{total}

Vol.
NaOH
Added [mL] |
Total
Volume
[mL] |
% Diss
Acid |
n_{HA} |
n_{A}- |
n_{[H+]e}
mmol |
n_{[OH-]e}
mmol |
pH |

0 | 25 | 1.0 | 2.5 | 0 | 2.488 x 10^{-2} |
2.513 x 10^{-10} |
3.00 |

10 | 35 | 40 | 1.5 | 1.0 | 5.245 x 10^{-4} |
2.335 x 10^{-8} |
4.82 |

12.5 | 37.5 | 50 | 1.25 | 1.25 | 3.748 x 10^{-4} |
3.777 x 10^{-7} |
pK_{a} = 5.00 |

20.0 | 45.0 | 80 | 0.500 | 2.000 | 1.125 x 10^{-4} |
1.801 x 10^{-7} |
5.602 |

22.0 | 47.0 | 88 | 0.300 | 2.200 | 6.408 x 10^{-5} |
3.448 x 10^{-7} |
5.865 |

24.0 | 49.0 | 96 | 0.100 | 2.400 | 2.041 x 10^{-5} |
1.176 x 10^{-6} |
6.380 |

25 | 50 | 100 | 3.54 x 10^{-4} |
2.5 | 7.07 x 10^{-8} |
3.54 x 10^{-4} |
8.85 |

26 | 51 | 100 | 2.5 x 10^{-9} |
2.5 | 2.60 x 10^{-10} |
0.1 | 11.29 |

28 | 53 | 100 | 8.3 x 10^{-9} |
2.5 | 9.363 x 10^{-11} |
0.3 | 11.75 |

30 | 55 | 100 | 5.0 x 10^{-9} |
2.5 | 6.050 x 10^{-11} |
0.5 | 11.96 |

40 | 65 | 100 | 1.7 x 10^{-9} |
2.5 | 2.82 x 10^{-11} |
1.5 | 12.36 |

50 | 75 | 100 | 1.0 x 10^{-9} |
2.5 | 2.25 x 10^{-11} |
2.5 | 12.52 |

The shape of the titration curve for a weak acid with a strong base will depend on the value of its' Ka.

The Titration of a Weak Acid with a Strong Base is calculated in 4 steps:

- Calculate the pH of the pure acid before the titration starts
- Calculate the pH of the partially neutralized acid as if it were a buffer for additions of base between zero and the equivalence point. (Note: The Herderson-Hasselbach eqn. fails for small concentrations of relatively strong acids)
- Calculate the pH at the equivalence point as if the solution were a pure solution of the salt (conjugate base) of the acid
- Calculate the pH of the solution beyond the equivalence point as if it were a pure solution of the excess base

See below for 25 mL of 0.100 M acid (HA) titrated with 0.100 M NaOH (with three different values of the pKa of HA):

**What does the titration curve for a weak base titrated with a strong acid
look like?**