__Standard solutions__

__The standard solutions used in neutralization titrations are strong
acids or bases for complete reaction and sharper end points.__

- Usual acids are HCl, H
_{2}SO_{4}and HClO_{4}used as cold dilute solutions for safety. HNO_{3}is too oxidizing and gives unwanted side reactions. - Usual bases are NaOH, KOH and Ba(OH)
_{2}.

__An acid-base INDICATOR is a weak organic acid or weak organic base that
shows a color change between the acid or base and the conjugate form.__

_{
}

or

_{
}

If we just consider the acid form, we see from a definition of the
dissociation constant K_{a}, that the color change is determined by the
hydronium ion concentration.

_{
}

The range of the indicator, from what the human eye distinguishes as completely acid to completely base color change, is about

_{
}

or about a two orders of magnitude change in the indicator concentration ratio from

_{
}

Example: An acidic indicator with a dissociation constant of
10^{-5} and exhibiting a color change from pink (acid) to yellow
(conjugate base) has a pK_{a} = 5 and therefore will be pink at about
pH = 4 and yellow at pH = 6.

**Titration curve for a strong acid and strong base.**

__To calculate a titration curve for a strong acid with a strong base,
calculations must be made before, at and after the equivalence point
(preequivalence, equivalence and postequivalence).__

In the preequivalence stage, we can calculate the concentration of analyte
(acid) from the starting concentration and the volumetric data (amount of
titrant added). At the equivalence point the H_{3}O^{+} and
OH^{-} concentrations are equal and the concentrations can be derived
from the ion-product constant of water, K_{w}. Postequivalence, the
excess OH^{-} concentration can be computed from the volumetric data.

A useful expression for pOH and pH comes from the ion-product constant for water

_{
}

Example: Calculate the titration curve for 50 mL of 0.1 M HCl titrated with 0.1 M NaOH.

Since HCl is a strong acid,

_{
}

After addition of 10 mL 0.1 M NaOH, the total solution volume increases to 60 mL and the concentration of the acid is decreased,

_{
}

Other preequivalence points can be determined in a similar way.

At the equivalence point, the H_{3}O^{+} and
OH^{–} concentrations are equal and we can use the ion-product for
water to calculate the pH,

_{
}

At postequivalence, the solution now contains an excess of
OH^{–}, and we can switch to calculating [OH^{–}]. For
example after addition of 60 mL 0.1 M NaOH,

_{
}

and pOH is

_{
}

and

_{
}

The data from several calculations for different NaOH additions produces a sigmoidal titration curve. Titration curves of a strong base with a strong acid can be derived in a similar way to that above.

The effect of reagent concentration is shown in the following figure.

(Data from Table 12-2, page 270 in Skoog, West, Holler and Crouch)

__Choice of indicator__

__This figure also illustrates that the choice of indicator is not
particularly crucial for the more concentrated reagents because a large change
in pH occurs around the equivalence point. Any indicator with a color change
in the pH range 4-10 will perform about equally. However, if the same choice
of indicators is used for the more dilute reagents, a problem is obvious. Any
indicator which begins to change color around pH 4.5 will produce an incorrect
endpoint determination. Similarly, an indicator with a color change around pH
9 will produce a significant error in the end point determination. Only an
indicator which changes color close to the true equivalence point (in this case, pH 7) will
give a precise end point determination.
__

__Buffer solutions__

__A BUFFER SOLUTION is a a conjugate acid-base or conjugate base-acid
solution that resists a change in pH when additional acid or base is added, or when diluted.
__

A buffer solution occurs whenever a weak acid is titrated with a strong base or vice versa. Therefore, before we can investigate such titrations we need to know something about buffer solutions.

Buffers are used throughout chemistry whenever a constant pH must be maintained.

__Weak acid-conjugate base buffers__

__A solution containing a weak acid, AcidH, and its conjugate base,
Acid__^{-}, may be acidic, neutral or basic depending upon the relative
magnitudes of two competing reactions,

_{
}

If the equilibrium of the FIRST equation lies farther to the RIGHT than the second, the solution is ACIDIC. If the SECOND equilibrium lies further to the RIGHT than the first, the solution is BASIC.

The two equations above show that the hydronium and hydroxide ion
concentration is dependent upon both K_{a} and K_{b} and also
upon the concentration of the acid and its conjugate base. In order to
calculate the pH of the solution we will need to know the analaytical
concentration of acid [AcidH] and the conjugate base [Acid^{-}].

An examination of the two equations above shows that in the first, the [AcidH]
decreases by an amount corresponding to an increase in
[H_{3}O^{+}]. In the second, the [AcidH] increases by an
amount corresponding to an increase in [OH^{-}]. Therefore, the
concentration of the AcidH is related to its analytical concentration by,

_{
}

A similar idea applies to calculating the concentration of the
conjugate base. The first equation will produce an increase in
[Acid^{-}] corresponding to an increase of
[H_{3}O^{+}]. The second equation produces a decrease in
[Acid^{-}] corresponding to an increase in [OH^{-}].
Therefore,

_{
}

Often, the concentration of hydronium and hydroxide is very much small than the analytical concentrations of the acid and conjugate base. The above two equations then simplify to,

_{
}

Substituting these into the dissociation constant expression gives,

_{
}

The last three expressions break down if the analytical concentrations
of the acid and conjugate base are very small or when the dissociation constant
of the acid or base are very large (>=10^{-3}).

Example: What is the pH of a solution that is 0.5 M in benzoic acid
and 1.00 M in sodium benzoate? The dissociation constant of the acid is
6.28x10^{-5} (from Appendix 2 in Skoog, West, Holler and Crouch).

The equation is

_{
}

and since we can assume

_{
}

we can write

_{
}

A similar approach can be used for weak bases with conjugate acids.

Example: Calculate the pH of a solution that is 0.05 M in
NH_{3} and 0.075 M in NH_{4}Cl.

The appropriate equations are

_{
}

_{
}

By consideration of the above equations, we see that the first
reaction decreases the concentration of NH_{3} by an amount equal to
[OH^{-}] and the second increases the [NH_{3}] by an amount
equal to [H_{3}O^{+}]. Or,

_{
}

and the first equation increases the [NH_{4}^{+}] by
an amount equal to [OH^{-}] and the second decreases the
[NH_{4}^{+}] by an amount equal to [H_{3}O^{+}]
or

_{
}

A consideration of the K_{a} and K_{b} values above
suggests that K_{b}>>K_{a}. So we can write,

_{
}

Since the concentration of OH^{-} is very much smaller than either of
the analytical concentrations, [NH_{3}]~0.05 and
[NH_{4}^{+}]~0.075 M.

Now, from the definition of K_{a} we can obtain the pH of the
solution.

_{
}

and pH=-log(8.55x10^{-10})=9.07. If we now calculate the
[OH^{-}] we can check the validity of our assumption that [OH-] is very
much smaller than the analytical concentrations.

_{
}

__Properties of buffers__

* Buffers maintain pH at an almost constant level for small
additions of acid and base. This effect can be calculated in a similar way to
the examples above.

Example: In the case of the system of the previous example, what is the pH change upon addition of 10 mL of 0.1 M NaOH to 500 mL of the buffer solution.

The addition of NaOH converts part of the buffer to
NH_{4}^{+}.

_{
}

The analytical concentrations of the NH_{3} and
NH_{4}^{+} then become,

_{
}

When inserted into the acid dissociation constant expression

_{
}

which corresponds to a pH of -log(8.05x10^{-10})=9.09 compared
to the original pH of 9.07.

* They also maintain an approximately constant pH with dilution
(until the [OH^{-}] or [H_{3}O^{+}] become comparable
with the analytical concentrations of the reagents.

__Buffer capacity__

__The BUFFER CAPACITY of a solution is the number of moles of strong acid
or base that 1 L of buffer can absorb before causing a unit change in pH. It depends upon
both the CONCENTRATION of the acid/base and on their CONCENTRATION RATIO.
Buffer capacity is maximum when the concentration ratio of acid and conjugate
base is unity.
__

__Preparation of buffers__

__In theory it is possible to prepare a buffer of almost any pH. However,
uncertainties in the dissociation constants and activity coefficients of the
solution means that the pH of a buffer can only be calculated approximately.
In practice, a buffer of a desired pH is prepared by an approximate method and
the pH adjusted to exactly that desired by addition of small quantities of
strong acid or base.
__

**Titration curves for weak acids or weak bases with strong acids or
bases**

__Here we need four types of calculation:__

- At the beginning, the solution is entirely composed of the weak acid or
base and the pH can be calculated from its concentration and dissociation
constant.
- After addition of a small quantities of base (or acid), the solution is
essentially a series of buffers anf the pH can be calculated from the
analytical concentration of the conjugate base and the residual concentration
of the weak acid.
- At the equivalence point, the solution contains only the conjugate of the
weak acid or base being titrated and the pH can be calculated from this
product.
- Beyond the equivalence point, the pH is largely dominated by the excess
strong acid or base added.

_{
}

(1) Using the expression derived before,

_{
}

so pH = –log(1.15x10^{-3}) = 2.94.

(2) When 10 mL of KOH is added, a buffer solution of AcOH and KOAc has been formed.

_{
}

Substituting this into the expression for the dissociation constant,

_{
}

and pH=-log(2.03x10^{-5})=4.69.

(3) At the equivalence point, (18.75 mL KOH) all the acetic acid has been converted to potassium acetate. The pH is therefore similar to dissolving potassium acetate in water,

_{
}

Neglecting water dissociation

_{
}

And, the sum of all sources of acetate must equal the analytical concentration of the acetate,

_{
}

Substituting these into the expression for K_{b},

_{
}

Here we have assumed that
[OH^{–}]<<*c*_{OAc}-.

Therefore,

_{
}

And pH is

_{
}

(4) After addition of extra base, say 19.00 mL of KOH, the pH is
largely determined by the excess strong base. As a first approximation, the
OH^{–} produced by the acetate ion is suppressed by the strong base and
so we need only consider the OH^{–} produced by the base,

_{
}

and

_{
}

Further calculations will complete the titration curve.

__Half Equivalence Point__

__If we repeat the buffer calculation for a weak acid ((2) above)
for a volume of KOH which is the equivalence point divided by 2 (18.75 mL/2) we
notice that at the HALF EQUIVALENCE POINT the concentrations of
H___{3}O^{+} and OH^{-} are the same. As such, their
concentrations cancel in the expression for K_{a}, and the
[H_{3}O^{+}]=K_{a}.

(2) When 9.375 mL of KOH is added, a buffer solution of AcOH and KOAc has been formed.

_{
}

Substituting this into the expression for the dissociation constant,

_{
}

Similarly, in the titration of a weak base, the [OH^{-}] at the
HEP is equal to K_{b}.

The BUFFERING CAPACITY is maximal at the HEP. Hence, the effect on the titration curves of varying concentration around this point is very small.

A titration curve for a similar concentration of strong acid would be
much 'flatter' in the region up to the equivalence point. The magnitude of the
pH change about the equivalence point decreases as K_{a} becomes
smaller, making endpoint detection much more difficult. (See Figure 12-10 in
Skoog, West, Holler and Crouch).

__Indicator Choice for Minimization of Titration Error.__

__The difference between the equivalence point and the end point (the
titration error) should be minimal if accurate results are to be obtained. The
choice of indicator is critical in this process as illustrated above.
__

Clearly, Bromothymol green is an unsuitable indicator for this reaction. It would suggest an endpoint of about 20 mL instead of the true 50 mL. Bromothymol blue is a better choice, but the color change from yelow-green to blue would indicate an endpoint some 2-3 mL too small. Phenolphthalein is the correct indicator for this reaction since the endpoint would be signalled at ~50 mL.