### Richard A. Paselk

Chem 109

General Chemistry

Spring 2009

Lecture Notes:: 27 April

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# The pH Scale

The concentration of hydronium ion in water is extremely influential on all kinds of chemistry. The range of hydronium ion concentration in water is also vast, with extremes of about 10M to about 10-15M, and commonly ranging from 1M - 10-14M. Imagine plotting [H3O+] vs. volume of acid added to a base solution in a titration from 10-14M - 1M. If you had one cm on the graph paper = 10-14M, then you would need a piece of paper 109 km long (greater than the distance from the Sun to Jupiter) to plot this titration! Obviously a more convenient measure is needed. This is easily accomplished by looking instead at the logarithm of [H+] and defining a new term,

pH = -log[H+]

Because of the equilibrium dissociation of water to H+ + OH-, the concentration of hydrogen ion in water is related to the concentration of hydroxide ion:

H2O H+ + OH-, so

K = [H+][OH-] / [H2O]

But the concentration of water remains essentially the same in dilute solution, so by convention we define the dissociation constant or ion product for water:

Kw= [H+][OH-] = 1.0 x 10-14 @ 25 °C

Let's look at some general characteristics of pH in aqueous solution.

• Range: pH = -1 to pH = 15 (10M -10-15M)
• for 1 M HCl, pH = 0
• for 1 M NaOH, pH = 14
• At midrange [H+] = [OH-] = 10-7M. The solution is said to be "neutral."
• This follows in aqueous solution from Kw = 1.0 x 10-14 = [H+] [OH-], thus if [H+] = [OH-], then [H+] = (1.0 x 10-14)1/2= 1.0 x 10-7
• Low pH means acidic:
• For 1M strong acid, pH = 0.0 (log 1 = 0)
• For 0.1M strong acid, pH = 1.0
• For 10-7M H+, pH = 7
• High pH means basic:
• For 1M strong base, pH = 14 ([H+] = (1.0 x 10-14) / [OH-] = (1.0 x 10-14) / 1 = 1.0 x 10-14 and pH = -log(1.0 x 10-14) = 14.0.
• For 0.1 M OH-, (1.0 x 10-14) / 0.1 = 1.0 x 10-13 and pH = -log(1.0 x 10-13) = 13.0.
• For 10-7M OH-, (1.0 x 10-14) / (10-7) = 1.0 x 10-7 and pH = -log(1.0 x 10-7) = 7

Examples:

• What is the pH of a solution of 0.015 M HCl?
Strong acid, so [H+] = 0.015 M
pH = - log [H+]
pH = - log 0.015 = - (- 1.824)
pH = 1.82

Note that the significant figures are correct, 1 is the power of ten, only the figures to the right are significant.

• What is the pH of a solution of 0.067 M NaOH
Strong base, so [OH-] = 0.067 M
Recall that [H+][OH-] = 1.0 x 10-14
Substituting, [H+][0.067] = 1.0 x 10-14
Rearranging, [H+] = (1.0 x 10-14) / 0.067 = 1.493 x 10-13
pH = - log (1.493 x 10-13) = - (- 12.83)
pH = 12.83
Again note the significant figures - 12 corresponds to the power of ten, only the figures to the right are significant.

Note that the "p" has the more general meaning of "-log[]". Thus pOH is -log [OH-], pCa = -log [Ca2+], etc.

## pH of weak acid solutions

Weak acid dissociations involve equilibria. The equilibrium constants have a specific symbol = Ka.

Example: What is the pH of a 0.10 M solution of acetic acid. Ka = 1.8 x 10-5

 HOAc H+ + OAc- Before reaction 0.10 M 0 0 @ Equilibrium 0.10 M- x x x

Ka = [H+][OAc-] / [HOAc]

assume x << 0.1 since Ka =1.8 x 10-5, then [HOAc] = 0.10 M

Substituting, Ka = (x)(x) / 0.10 = 1.8 x 10-5,

x2 = 1.8 x 10-6

x = 1.34 x 10-3M; assumption OK.

pH = - log (1.34 x 10-3) = 2.87

Notice the significant figures. For a log function the number in front of the decimal is the exponent of ten, thus pH = 2.87 is a 2 significant figure number.

# Acid Equilibria

Buffer calculations: One of the most frequent calls for calculating acid equilibria is calculations involving buffers. What is a buffer?

• A buffer is a solution which resists changes in pH. Essentially it consists of an acid and its salt (an acid and its conjugate base) in solution together. Thus the solution has a proton donor and a proton acceptor, so pH is stabilized.
• A buffer is simply an acid equilibrium system with significant amounts of both the acid and its conjugate base.

With this in mind let's do some examples.

Example: Calculate the pH of a buffer made up by dissolving 0.0125 moles acetic acid (HOAc) and 0.0250 moles of sodium acetate (NaOAc) in enough water to make 1.000 L of solution. Ka = 1.8 x 10-5

 HOAc H+ + OAc- Before reaction 0.0125 moles/L 0 0.0250 moles/L @ Equilibrium (0.0125 - x) M assume x is small, = 0.0125 x (0.0250 + x) M assume x is small, = 0.0250

Ka = [H+][OAc-] / [HOAc]

Substituting, Ka = [H+](0.0250) / (0.0125) = 1.8 x 10-5

Rearranging, [H+] = (1.8 x 10-5)(0.0125) / (0.0250) = 0.90 x 10-5

x is within experimental error (0.000009 < ±0.0001), so assumption OK

pH = 5.046

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