pH
Copyright, 1992 F.W. Boyle, Jr. PhD.
Just what does the little "p" mean.
A number of terms pH, pOH, pKa, and pKb have been introduced.
One main point to remember is that the small "p" simply means:
-log
That is "p" before H or OH means the negative of the log of
the hydrogen or hydroxide ion concentrations.
How to calculate pH
First we must understand logarithms. The term, "log",
indicates that the conversion is for a base of 10. That is, we
are trying to find the power of 10 that will give us the number
that we are taking the log of. For example:
log(2) = 0.3010
and
100.3010 = 2
Some other points to remember are:
log(4) = 0.6020
and
log(4) = log(22) = 2(log(2))
On to pH
Given that [H+] = 8.13 x 10-4, calculate the pH of the
solution.
1. Fast and simple.
Enter 8.13 then press the ee or scientific button and enter
the -4. Next press the LOG button. Then press the +/- button to
change the sign. You have taken the log and calculated the pH.
2. By definition the log of 10 to any power is equal to that
power, i.e.
log(10-4) = -4
and the log of two numbers multiplied together is equal to the
sum of the logs of the individual numbers, e.g.
log(8.13 x 10-4) = log(8.13) + log(10-4)
since we know the answer to the second part is -4, all that needs
to be done is to enter 8.13 and take the log.
log(8.13) = 0.91
therefore,
log(8.13 x 10-4) = 0.91 + (-4)
= -3.09
This simplifies the information which must be entered.
Calculating [H+] from the pH.
A solution has a pH of 4.15. What is the [H+] ?
1. Fast and simple.
pH = -log[H+] = 4.15
-log[H+] = 4.15
Multiply by -1:
log[H+] = -4.15
Now raise both sides as powers of 10. (This is the same as
taking the antilog).
10log[H+] = 10-4.15
The lefthand side reduces to [H+] since 10log(x) is equal to
x. The 10 and log cancel each other just as was shown above.
Therefore the answer is:
[H+] = 10-4.15 M
The answer can also be written 7.08 x 10-5 M.
Calculating pH, [H+], and [H3O+]
Consider a 0.050 M solution of CH3COOH. In order to
calculate the pH, we must first calculate the [H+] or [H3O+]. We
know the [H+] = [H3O+] so we can use either one.
Writing the reaction:
CH3COOH(aq) <---> H+(aq) + CH3COO-(aq) Ka = 10-4.76 M
Ka = [H+] [CH3COO-]
[CH3COOH]
Initial conditions are:
0.050 M CH3COOH, 0 M H+, and 0 M CH3COO-
At equilibrium:
(0.050 - x) M CH3COOH, x M H+, and x M CH3COO-
Substituting:
Ka = (x M) (x M) = 10-4.76 = 1.74 x 10-5 M
(0.050 - x M)
.................................
Solving the equation:
x2 = (1.74 x 10-5 M) (0.050 - x M)
x2 + 1.74 x 10-4(x) + 8.7 x 10-7 = 0
_________________________________
x = -1.74 x 10-5 +/- [((1.74 x 105)2 - 4(1)(8.7 x 10-7]1/2
2(1)
x = [H+] = 9.24 x 10-4 M
Rather than solving for the two roots by using the quadratic
formula. The answer can be found by the method of successive
approximations as follows.
Writing the reaction:
CH3COOH(aq) <---> H+(aq) + CH3COO-(aq) Ka = 10-4.76 M
Ka = [H+] [CH3COO-]
[CH3COOH]
Initial conditions are:
0.050 M CH3COOH, 0 M H+, and 0 M CH3COO-
At equilibrium:
(0.050 - x) M CH3COOH, x M H+, and x M CH3COO-
Substituting:
Ka = (x M) (x M) = 10-4.76 = 1.74 x 10-5 M
(0.050 - x M)
.................................
For the first approximation, we assume x is much smaller than
0.050 so the denominator can be approximated by 0.050 rather than
0.050 - x. Substituting:
Ka = (x M) (x M) = 1.74 x 10-5 M
(0.050 M)
..................
Solving for x we get:
x2 = 1.74 x 10-5 M (0.050 M)
= 8.7 x 10-7 M2
x = [H+] = 9.33 x 10-4 M
Now we substitute this first approximation of x into the
original equation as follows:
Ka = (x M) (x M) = 1.74 x 10-5 M
(0.050 - 9.33 x 10-4 M)
And solve for x again:
x2 = 1.74 x 10-5 M (0.050 - 9.33 x 10-4 M)
= 8.54 x 10-7 M2
x = [H+] = 9.24 x 10-4 M
A third repeat will result in the same answer being
calculated. Therefore the answer by approximation is:
[H+] = 9.24 x 10-4 M