Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 109 - General Chemistry - Spring 2011

Lecture Notes 39: 29 April

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The pH Scale, cont.

Recall

pH = -log[H+]

and

Examples:

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  equilibrium arrow 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?

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  equilibrium arrow 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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© R A Paselk

Last modified 27 April 2011