Calculation of [HI from pH
345
2.00 x 10~
4 moles
l °
H -
] =
0.0760 liter
=
2 '
63 X I0
"
M
pOH = -log (2.63 x 10:)
) = -log 2.63 - log (lO"
3 )
= -0.42 + 3.00 = 2.58
pH = 14.00 - 2.58 = 11.42
CALCULATION OF LH
+ ] FROM pH
The preceding problems illustrate the computation of the pH for acidic,
basic, and neutral solutions. We also need to understand the reverse calculation—how to go from the pH or pOH to the actual concentration of acid
or base in a solution.
PROBLEM:
What is the hydrogen-ion concentration in a solution whose pH is 4.30?
SOLUTION:
From the definition of pH,
[H
+ J = antilog (-pH) = IQ1 -
30
If you simplify this with a hand calculator, enter -4.3 through the keyboard
(enter 4.3 followed by the change-sign key), then proceed as on p 16. This
gives [H
+ ] = 5.01 x 10~
5 M.
If you simplify by using a log table, then the decimal part of this exponent
(-4.30) must first be changed to a positive number. To do this, rewrite it as
10°70 x l(r
5 . The antilog of 0.70 (from the log table) is 5.01. This gives
[H
+ ] = 5.01 x ID"
5 M
PROBLEM:
What is the hydroxide-ion concentration in a solution whose pH is 8.40?
SOLUTION:
Because we want [OH~], we first convert to pOH:
pOH = 14.00 - pH = 14.00 - 8.40 = 5.60
[OH-1 = antilog (-pOH) = 10S -
B "
If you simplify this with a hand calculator, enter -5.6 through the keyboard
(enter 5.6 followed by the change-sign key), then proceed as on p 16. This
gives [OH-J = 2.51 x IQ-" M.
If you simplify by using a log table, then the decimal part of this exponent