Chem 109 - General Chemistry - Spring 2011

Lecture Notes 39: 29 April

The pH Scale, cont.

Recall

pH = -log[H⁺]

and

Range: pH = -1 to pH = 15 (10M -10^-15M)
At midrange [H⁺] = [OH^-] = 10^-7M. The solution is said to be "neutral."
- This follows in aqueous solution from K_w = 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 = K_a.

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

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

K_a = [H⁺][OAc^-] / [HOAc]

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

Substituting, K_a = (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. K_a = 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