### Richard A. Paselk

Chem 107

Fundamentals of Chemistry

Fall 2009

Lecture Notes: pH and Acid/Base Examples

Worked 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
Note that the significant figures are correct, 12 is the power of ten, only the figures to the right are significant.

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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