Cracking the SAT Chemistry Subject Test

KOH(aq) →   K+(aq)  +   OH−(aq)

Second,  realize     that    there   is  really  no  KOH(aq)     in

solution.   It’s    all dissociated.    Therefore,  what    we  really

have    is

KOH(aq) →   K+(aq)  +   OH−(aq)

1.0M                                                1.0 M                               1.0 M

Third,  take    the pOH since   that’s  what    we  have.

pOH =   −log    [OH−]

                                =   −log    (1.0    M)

                                =   −log    (10^0 M)

                                =   0

Fourth, recall  that    pH  +   pOH =   14  (at 25°C),  and solve

for pH.

pH  =   14  −   pOH

                        =   14  −   0   =   14

It is also a good idea to remember that for 1.0 M strong base, pH = 14 because
these solutions are also commonly used in the SAT Chemistry Subject Test
laboratory questions.

Weak Acids and Bases

Acids and bases that partially, reversibly dissociate are referred to as weak acids
or bases. Again, the term weak is NOT used as a common adjective in acid-base
chemistry. It has a very specific meaning; it means partial or reversible
dissociation. For example, HF is a weak acid, and NH 3 is a weak base.

HF(aq)      H+(aq)  +   F−(aq)

NH 3 (aq)   +   H 2 O(l)        NH 4 +(aq)  +   OH−(aq)