The Foundations of Chemistry

strongly to the left.As a result of the diminished OH
concentration, the reaction

NH 3 H 2 O 34 NH 4 OH
 (shifts right)

shifts markedly to the right.Because the [NH 3 ] in the buffer solution is high, this can
occur to a great extent. The net reaction is

H 3 O NH 3 88n NH 4 H 2 O (
100%)

added acid base weak acid water

When a strong base such as NaOH is added to the originalbuffer solution, it is neutral-
ized by the more acidic component, NH 4 Cl, or NH 4 , the conjugate acid of ammonia.

NH 3 H 2 O 34 NH 4 OH
 (shifts left)

Because the [NH 4 ] is high, this can occur to a great extent. The result is the neutral-
ization of OH
by NH 4 .

OH
NH 4 88nNH 3 H 2 O (
100%)

or, as a formula unit equation,

NaOH NH 4 Cl88n NH 3 H 2 ONaCl (
100%)

added base acid weak base water

Summary Changes in pH are minimized in buffer solutions because the basic

component can react with H 3 Oions and the acidic component can react with OH


ions.

PREPARATION OF BUFFER SOLUTIONS

Buffer solutions can be prepared by mixing other solutions. When solutions are mixed,
the volume in which each solute is contained increases, so solute concentrations change.
These changes in concentration must be considered. If the solutions are dilute, we may
assume that their volumes are additive.

EXAMPLE 19-5 Preparation of Buffer Solutions

Calculate the concentration of H 3 Oand the pH of a buffer solution prepared by mixing 200. mL
of 0.10 MNaF and 100. mL of 0.050 MHF. Ka7.2 10
^4 for HF.

Plan

Calculate the number of millimoles (or moles) of NaF and HF and then the molarity of each
solute in the solution after mixing. Write the appropriate equations for both NaF and HF,
represent the equilibrium concentrations algebraically, and substitute into the Kaexpression
for HF.

Solution

When two dilute solutions are mixed, we assume that their volumes are additive. The volume
of the new solution will be 300. mL. Mixing a solution of a weak acid with a solution of its
salt does not form any new species. So we have a straightforward buffer calculation. We calcu-
late the number of millimoles (or moles) of each compound and the molarities in the new
solution.

19-3

19-3 Preparation of Buffer Solutions 803

The net effect is to neutralize most of

the H 3 Ofrom HCl. This slightly

increases the ratio [NH 4 ]/[NH 3 ],

which governs the pH of the solution.

The net effect is to neutralize most of

the OH
from NaOH. This slightly

decreases the ratio [NH 4 ]/[NH 3 ],

which governs the pH of the solution.

Alternatively, one could substitute into

the Henderson–Hasselbalch equation

and solve for pH, then [H 3 O].