The Foundations of Chemistry

Solution

From the definition of pH, we write

log [H 3 O
]3.301

Multiplying through by 1 gives

log [H 3 O
]3.301

Taking the inverse logarithm (antilog) of both sides of the equation gives

[H 3 O
] 10 3.301 so [H 3 O
]5.00 10 ^4 M

You should now work Exercise 26.

A convenient relationship between pH and pOH in all dilute solutions at 25°Ccan be

easily derived. We start with the Kwexpression.

[H 3 O
][OH]1.0 10 ^14

Taking the logarithm of both sides of this equation gives

log [H 3 O
]
log [OH]log (1.0 10 ^14 )

Multiplying both sides of this equation by 1 gives

(log [H 3 O
])
(log [OH])log (1.0 10 ^14 )

or

pH
pOH14.00

We can now relate [H 3 O
] and [OH] as well as pH and pOH.

[H 3 O
][OH]1.0 10 ^14 and pH
pOH14.00 (at 25°C)

From this relationship, we see that pH and pOH can bothbe positive only if bothare less

than 14. If either pH or pOH is greater than 14, the other is obviously negative.

Please study carefully the following summary. It will be helpful.

At any temperature, pH
pOH
pKw.

Remember these relationships!

758 CHAPTER 18: Ionic Equilibria I: Acids and Bases

Solution General Condition At 25°C

acidic [H 3 O
][OH][H 3 O
]1.0 10 ^7 M[OH]

pHpOH pH 7.00 pOH

neutral [H 3 O
][OH][H 3 O
]1.0 10 ^7 M[OH]

pHpOH pH 7.00 pOH

basic [H 3 O
][OH][H 3 O
]1.0 10 ^7 M[OH]

pHpOH pH 7.00 pOH

See the Saunders Interactive
General Chemistry CD-ROM,
Screen 17.4, The pH Scale.

To develop familiarity with the pH and pOH scales, consider a series of solutions

in which [H 3 O
] varies from 10 Mto 1.0 10 ^15 M. Obviously, [OH] will vary from

1.0 10 ^15 Mto 10 Min these solutions. Table 18-3 summarizes these scales.