THE pH SCALE AND LOGARITHMS
Chemistry NAME:___________
Unit: Acids and Bases GIA 10-3 Reading: Chapter 19
Working Mode: Pairs. Solve the problems on a
separate sheet of notebook paper. Each
student must submit his own work. Include
the name of your partner.
Learning Objective: Calculate between [H3O+]
and pH, [OH-] and pOH for a variety of types of solutions.
Background:
The concentration of the hydronium ion (H3O+) in an
acidic or basic solution is often indicated by the “pH.”
The term “pH” is actually a shorthand version of the expression,
-log10[H3O+]. Therefore it is necessary for Chemistry students to be able
to operate with the log function.
Logarithms are a function that can be used to determine the
value of an exponent in an expression, such as
y = bx . In
solving for “x” one would need a logarithmic function.
The exponential function, y = bx, could be rewritten as x =
logb y to solve for “x.”
Part 1 –
Calculating pH and [H3O+]: In Chemistry the function
is found in the study of acids and bases. For
a particular acid solution with an H3O+ concentration of
1.0x10-2 M, the pH of this solution is 2, since –log (1.0 x 10-2)
= -(0-2) or 2.
Try these on your calculator:
(a)
[H3O+] = 1.0
x 10-5;
pH =
(b)
[H3O+] = 1.0
x 10-11;
pH =
(c)
[H3O+] = 6.5
x 10-3;
pH =
Now if the pH is given, how does one find the [H3O+]? The relationship between the [H3O+] and pH looks like this when calculating [H3O+] from pH: [H3O+] = 10-pH. So if the pH of a particular acid solution is 8.0 then the [H3O+] = 10-8 or 1.0 x 10-8 M. If the pH is not a whole number then the value of the coefficient will not be 1.0. i.e. pH = 4.50, then [H3O+] = 10-4.50 = 3.2 x 10-5 M. (Note you may NOT leave your answer in terms of 10-4.50, you must give the concentration in correct exponential notation. Depending on your calculator you may have a 10x button or you may need to type the exponent value in and they press the “INV” (for inverse) and “LOG” keys in sequence.
Try these on your calculator:
(a) pH = 3.0; [H3O+] =
(b) pH = 4.90; [H3O+] =
(c) pH = 12.76; [H3O+] =
(d) pH = .40; [H3O+] =
Part 2 –
Calculating between [H3O+], [OH-], pH and pOH:
Finally, since the “p” in pH is shorthand for “-log10” one
can operate with this “p” on other concentrations or numbers such as
“pOH” for –log [OH-] or the “pKa“ for –log Ka,
where Ka is the equilibrium value for the acid dissociation constant.
Furthermore, since the self-ionization constant expression for water is Kw
= [H3O+][OH-] = 1.0 x 10-14 (at 25oC),
we can relate the [H3O+] to [OH-].
And since we can take the –log of any of these numbers, then pKw
= pH + pOH = 14.
Solve these:
[H3O+] = 4.7 x 10-3 M pOH =
[OH-] = 3.5 x 10-4 M pH =
pH = 9.75 [OH-] =
pOH = 6.41
[H3O+] =
Part 3 – Determining pH of acidic or basic solutions: For strong acids and bases, determining the pH of the solution is a matter of converting the given acid or base solute concentration and determining the key dissociated ion’s concentration. (This part, by the way, is similar to the third computer-based, Mastering Chemistry exercise.)
Give the pH for the following solutions:
0.0100 M NaOH
pH =
1.0 x 10-5 M HBr
pH =
0.00675 M KOH
pH =
0.542 M HNO3
pH =
0.544 M H2SO4 (both hydrogens ionize)
pH =
8.72 x 10-3 M Ca(OH)2 pH =
