Ch107_Lec_pH_Acid_Base Exercises

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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*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
	Use Back Button to Return to Notes

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. K_a = 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 K_a is very small (1.8 x 10^-5)

@ Equilibrium

HOAc = 0.10 M
x x

K_a = [H⁺][OAc^-] / [HOAc]

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!

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. 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

K_a = [H⁺][OAc^-] / [HOAc]

Substituting, K_a = [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 = pK_a + 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. pK_a = 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 = pK_a + 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.? pK_a = 4.74

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

pH = pK_a + log[A^-] / [HA]

Rearranging, log[A^-] / [HA] = pH - pK_a

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

Now get rid of logs by raising both sides to power of ten: 10^{log[A^-] / [HA]} = 10^0.51

[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? pK_a = 4.74.

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

pH = pK_a + log[A^-] / [HA]

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

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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© R A Paselk

Last modified 9 December 2005

Humboldt State University ® Department of Chemistry

Richard A. Paselk