Cracking the SAT Chemistry Subject Test

x   =   1   ×   10−2.5

Fifth,   don’t   worry   about   the     fractional  exponent

because it  will    disappear   when    you calculate   pOH.

pOH =   −log    [OH−]

                                =   −log    (1  ×   10−2.5 M)

                                =   2.5

Sixth,  remember    that    pH  +   pOH =   14  (at 25°C),  and

solve   for pH.

pH  =   14  −   pOH

                    =   14  −   2.5

                    =   11.5

Na+ and NaOH

Na+(aq) can also    be

thought of  as  Na(H 2 O)+,

or  simply  remember    that

aqueous metal   ions    are the

conjugates  of  their   metal

hydroxides.

Conjugate Acid/Base Pairs

A conjugate pair of molecules refers to two molecules that have identical
molecular formulas except that one of them has an additional H+.

Some examples of conjugate pairs are

HCl and Cl−

H 2 O   and OH−

H 2 PO 4 −  and HPO 4 2−

Na+ and NaOH

Some molecules/ions that are often mistaken for conjugate pairs are

H 3 O+/OH−, H 2 SO 4 /SO 4 2−,  H 2 CO 3 /CO 3 2−