Humboldt State University ® Department of Chemistry

Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: pH and Acid/Base Examples

© R. Paselk 2005
 
 

Worked Examples
 

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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
  0.10 M- x   x   x

Assume x << 0.10 since Ka is very small (1.8 x 10-5)

@ Equilibrium
HOAc = 0.10 M
  x   x

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

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!

 

 

 

 

 

Example. Calculate the pH of a "buffer" (a solution which resists changes in pH) 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

pH = 5.05

 

 

 

 

 

So now let's do the problem above using the Henderson-Hasselbalch equation, pH = pKa + log[A-] / [HA].

Calculate the pH of a "buffer" (a solution which resists changes in pH) 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. pKa = 4.74.. Notice that we are already past the "reaction" stage, so just set up for "@ Equilibrium."

  HOAc   H+ + OAc-
@ Equilibrium
(0.0125- x) M
assume x is small,
= 0.0125
  x  
(0.0250 - x) M
assume x is small,
= 0.0250

pH = pKa + log[A-] / [HA]

pH = 4.74 + log(0.250) / (0.125)

= 4.74 + 0.301

pH = 5.04

 

 

 

 

 

 

Example. What ratio of acetate ion to acetic acid would you need to make up an acetate buffer with a pH of 5.25.? pKa = 4.74

Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc-

pH = pKa + log[A-] / [HA]

Rearranging, log[A-] / [HA] = pH - pKa

Substituting, log[A-] / [HA] = 5.25 - 4.74 = 0.51

Now get rid of logs by raising both sides to power of ten: 10log[A-] / [HA] = 100.51

[A-] / [HA] = 3.236 = 3.2

[A-] = 3.2 [HA]

 

 

 

 

 

 

Example: How many moles of sodium acetate must be added to 0.120 moles of acetic and sufficient water to make a liter if we want a buffer with a pH of 4.50? pKa = 4.74.

Again use the Henderson-Hasselbalch equation to find the ratio of HOAc to OAc-

pH = pKa + log[A-] / [HA]

Rearranging and substituting, log[OAc-] / [HOAc] = pH - pKa

Substituting, log[OAc-] / [HOAc] = 4.50 - 4.74 = - 0.24

Raise both sides to power of 10: [OAc-] / [HOAc] = 0.575

[OAc-] =0.575 [HOAc]

[OAc-] = 0.069


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Last modified 9 December 2005