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188 Chemistry: Matter and Change
Solving Problems: A Chemistry Handbook
SOLVING PROBLEMS:
A CHEMISTRY HANDBOOK
CHAPTER 18
Use a log table or calculator to find that the antilog of Ϫ9.70 is
2.0 ϫ 10 Ϫ10 .
[H ϩ ] ϭ 2.0 ϫ 10 Ϫ10 M
To determine [OH Ϫ ], first use the pH value to calculate pOH.
pOH ϭ 14.00 Ϫ pH
pOH ϭ 14.00 Ϫ 9.70 ϭ 4.30
Now use the equation relating [OH Ϫ ] to pOH.
[OH Ϫ ] ϭ antilog (ϪpOH)
[OH Ϫ ] ϭ antilog (Ϫ4.30)
A log table or calculator shows that the antilog of Ϫ4.30 is
5.0 ϫ 10 Ϫ5 .
[OH Ϫ ] ϭ 5.0 ϫ 10 Ϫ5 M
As expected, [OH Ϫ ] Ͼ [H ϩ ] in this basic solution.
Practice Problems
8. The pH or pOH is given for three solutions. Calculate [H ϩ ] and
[OH Ϫ ] in each solution.
a. pH ϭ 2.80
b. pH ϭ 13.19
c. pOH ϭ 8.76
Calculating the pH of strong acid and strong base solutions
You learned in Section 18.2 that strong acids and strong bases ionize
completely when dissolved in water. This means that for strong
monoprotic acids, the concentration of the acid is the concentration
of the H ϩ ion because each acid molecule releases one H ϩ ion.
Similarly, for a strong base such as NaOH, the concentration of the
base equals the concentration of the OH Ϫ ion. However, some strong
bases contain two or more hydroxide ions in each formula unit. An
example is Mg(OH) 2 . For a solution of Mg(OH) 2 , [OH Ϫ ] is twice
the concentration of the base. For example, for a 3.0 ϫ 10 Ϫ5 M
Mg(OH) 2 solution, the concentration of OH Ϫ is 2(3.0 ϫ 10 Ϫ5 M) ϭ
6.0 ϫ 10 Ϫ5 M. As you learned earlier in this section, pH can be
calculated once [H ϩ ] or [OH Ϫ ] is known.
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