5 Steps to a 5 AP Chemistry

the two ions. The larger Kvalue predominates. If the larger value is Ka, the solution is

acidic. If the larger value is Kb, the solution is basic. In the rare case where the two values

are equal, the solution would be neutral.

The following table summarizes this information:

CATION FROM ANION FROM SOLUTION

Strong Base Strong Acid Neutral

Strong Base Weak Acid Basic

Weak Base Strong Acid Acidic

Weak Base Weak Acid Must be determined by comparing Kvalues

For example, suppose you are asked to determine if a solution of sodium carbonate,

Na 2 CO 3 , is acidic, basic, or neutral. Sodium carbonate is the salt of a strong base (NaOH)

and a weak acid (HCO 3 −). Salts of strong bases and weak acids are basic salts. As a basic salt,

we know the final answer must be basic (pH above 7).

Buffers

Buffersare solutions that resist a change in pH when an acid or base is added to them. The

most common type of buffer is a mixture of a weak acid and its conjugate base. The weak

acid will neutralize any base added, and the weak base of the buffer will neutralize any acid

added to the solution. The hydronium ion concentration of a buffer can be calculated using

an equation derived from the Kaexpression:

Taking the negative log of both sides yields the Henderson–Hasselbalch equation,

which can be used to calculate the pH of a buffer:

The weak base Kbexpression can also be used giving:

These equations allow us to calculate the pH or pOH of the buffer solution knowing

Kof the weak acid or base and the concentrations of the conjugate weak acid and its con-

jugate base. Also, if the desired pH is known, along with K, the ratio of base to acid can be

calculated. The more concentrated these species are, the more acid or base can be neutralized

and the less the change in buffer pH. This is a measure of the buffer capacity, the ability

to resist a change in pH.

Let’s calculate the pH of a buffer. What is the pH of a solution containing 2.00 mol of

ammonia and 3.00 mol of ammonium chloride in a volume of 1.00 L?

There are two ways to solve this problem.

Kb

NH 32 H O NH OH

=×

++

−

+−

181 10^5

4

.



[

[

]

log

[]

[

OH ] =

B]

[HB

and pOH = p

HB

b + b B]

+

− KK×+

pH = p

A

a HA]

K +

−

log

[]

[

[]

[

]

HO

HA]

3a[A

+=×

K −

Equilibrium  223