Barrons SAT Subject Test Chemistry, 13th Edition

Since the molar concentration of the acid times the volume of the acid gives
the number of moles of acid:

Ma × Va = moles of acid

and the molar concentration of the base times the volume of the base gives the
number of moles of base:

Mb × Vb = moles of base

then, substituting these products into the mole relationship, we get

MaVa = 2MbVb

Solving for Mb gives

Substituting values, we get

BUFFER SOLUTIONS

Buffer solutions are equilibrium systems that resist changes in acidity and
maintain constant pH when acids or bases are added to them. A typical laboratory
buffer can be prepared by mixing equal molar quantities of a weak acid such as
HC 2 H 3 O 2 and its salt, NaC 2 H 3 O 2. When a small amount of a strong base such as

NaOH is added to the buffer, the acetic acid reacts (and consumes) most of the

excess OH− ion. The OH− ion reacts with the H+ ion from the acetic acid, thus

reducing the H+ ion concentration in this equilibrium:

HC 2 H 3 O 2 (aq) H+(aq) + C 2 H 3 O 2 −(aq)

This reduction of H+ causes a shift to the right, forming additional C 2 H 3 O 2 −

ions and H+ ions. For practical purposes, each mole of OH− added consumes 1

mole of HC 2 H 3 O 2 and produces 1 mole of C 2 H 3 O 2 − ions.

When a strong acid such as HCl is added to the buffer, the H+ ions react with

the C 2 H 3 O 2 − ions of the salt and form more undissociated HC 2 H 3 O 2. This does