Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Spring 2008
Exercise: pH and Buffer Problems	© R. Paselk 2008
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pH and Buffer (Equilibrium) Problems

First need to ask, is this a strong or a weak acid? Sulfuric acid, hydrochloric acid and nitric acid are the three common strong acids we use in the lab, and which we have memorized.

However, sulfuric acid is a strong acid for the first dissociation only. As a result the acid concentration is the hydrogen ion concentration, so [H⁺] = 0.033 M

And we know that pH = - log [H⁺]

pH = - log 0.033 = - (- 1.481)

pH = 1.48

Note that the significant figures are correct, 1 is the power of ten, only the figures to the right are significant.

 

2. What is the pH of a solution of 0.055 M barium hydroxide (it completely dissociates at this concentration).

First write out the formula for Barium hydroxide = Ba(OH)₂

Note that the question states barium hydroxide is completely dissociated under these conditions, therefore it acts as a strong base.

As a result, [OH^-] = 2 (0.055) = 0.11 M

To find pH however we need to know the hydrogen ion concentration, [H⁺]. Of course we know hydrogen ion concentration is related to hydroxide concentration by the relationship,

[H⁺][OH^-] = 1.0 x 10^-14

Substituting, [H⁺][0.11] = 1.0 x 10^-14

Rearranging, [H⁺] = (1.0 x 10^-14) / 0.11 = 9.09 x 10^-14

pH = - log (9.09 x 10^-14) = - (- 13.04)

pH = 13.04

Note that the significant figures are correct, 13 is the power of ten, only the figures to the right are significant.

 

3. What is the pH of a 0.15 M solution of boric acid, H₃BO₃. K_a = 6.0 x 10^-10

As we saw with sulfuric acid, only the first dissociation will occur when the acid is dissolved in water, so K_a is for the first dissociation only:

	H3BO3		H+	+	H2BO3-
Before reaction	0.15 M		0		0
	0.15 M- x		x		x
Assume x << 0.15 since Ka is very small (6.0 x 10-10)
@ Equilibrium	H3BO3 = 0.15 M		x		x

K_a = [H⁺][H₂BO₃^-] / [H₃BO₃]

Substituting, K_a = (x)(x) / 0.15 = 6.0 x 10^-10,

x² = 9.0 x 10^-11

x = 9.487 x 10^-6 M; assumption OK.

pH = - log (9.487 x 10^-6) = 5.023 = 5.02

Notice the significant figures. For a log function the number in front of the decimal is the exponent of ten,

thus pH = 5.02 is a 2 significant figure number!

 

4. Calculate the pH of a buffer made up by dissolving 0.150 moles sodium dihydrogenphosphate (NaH₂PO₄) and 0.100 moles of sodium hydrogenphosphate (Na₂HPO₄) in enough water to make 0.500 L of solution. K_a = 6.3 x 10^-8 (FYI, this salt combination is commonly used in biological experiments, since it is a very effective buffer around pH 7.)

Note that sodium is a "spectator ion" and can be ignored in the buffer. Proceeding normally then, we first write the acid dissociation reaction for H₂PO₄^-:

	H2PO4-		H+	+	HPO4-2
Before reaction	0.150 moles/0.5L = 0.300M		0		0.100 moles/0.5L = 0.200M
@ Equilibrium	(0.300- x) M assume x is small, = 0.300M		x		(0.200 - x) M assume x is small, = 0.200M

K_a = [H⁺][HPO₄^-2] / [H₂PO₄^-]

Substituting, K_a = (x)(0.200) / (0.300) = 6.3 x 10^-8

Rearranging, [x] = (6.3 x 10^-8)(0.300) / (0.200) = 9.45 x 10^-8 ; assumption OK.

[H⁺] = 9.45 x 10^-8

pH = 7.02

 

Alternatively we could use the Henderson-Hasselbalch equation, pH = pK_a + log[A^-] / [HA]:

First write the acid dissociation reaction too determine HA and A^-:

HA H⁺ + A^-

H₂PO₄^- H⁺ + HPO₄^-2

 

Next find pK_a = -log K_a= -log (6.3 x 10^-8)

thus pK_a = 7.20

substituting

pH = 7.20 + log (.200)/(.300) = 7.20 + (-0.1761)

pH = 7.02

 

5. Calculate the pH of a buffer made up by dissolving 0.0230 moles ascorbic acid (H₂C₆H₆O₆) and 0.0440 moles of sodium ascorbate (NaHC₆H₆O₆) in enough water to make 0.250 L of solution. pK_a = 4.30.

	H2C6H6O6		H+	+	HC6H6O6-
Before reaction	0.0230 moles/0.25L = 0.0920M		0		0.0440 moles/0.25L = 0.176M
@ Equilibrium	(0.0920 - x) M assume x is small, = 0.0920M		x		(0.176 - x) M assume x is small, = 0.176M

Let HA = H₂C₆H₆O₆ and A^-= HC₆H₆O₆^-

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

pH = 4.30 + log(0.176M)/(0.0920M) = 4.30 + (0.2817)

pH = 4.58

 

6. What ratio of sodium dihydrogenphosphate (NaH₂PO₄) and sodium hydrogenphosphate (Na₂HPO₄) will give a buffer with a pH = 7.00? pK_a = 7.20

Easiest way to approach this is to use the Henderson-Hasselbalch equation to find the ratio of H₂PO₄^- to HPO₄^2-.

Let HA = H₂PO₄^-and A^-= HPO₄^2-

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

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

Substituting, log[A^-] / [HA] = 7.00 - 7.20 = -0.20

Now get rid of logs by raising both sides to power of ten: 10^{log[A-] / [HA]} = 10^-0.20

[A^-] / [HA] = 0.631 = 0.63

[A^-] = 0.63 [HA]

  7. Calculate the pH of the solution resulting when 0.250 L of 0.500 M acetic acid is mixed with 0.250 L of 0.350 M sodium hydroxide. pK_a = 4.74.

First we need to write out the reaction:

	CH2COOH	+	OH-		CH2COO-
Before reaction	(0.25L)(0.5M)/0.5L = 0.250M		(0.25L)(0.35M)/0.5L = 0.175M		0
1:1 reaction, so hydroxide is limiting and will be completely consumed.
@ Equilibrium	0.250 - 0.175 = 0.0750M		0		0.175M

Easiest to use the Henderson-Hasselbalch equation to find the pH

pH = pK_a + log[A^-] / [HA] = pK_a + log[CH₂COO^-] / [CH₂COOH]

Substituting, pH = 4.74 + log(0.175/0.0750) = 4.74 + (0.3679) =5.108

pH = 5.11

Check: Compare the concentrations of the acid and conjugate base.

• If the acid is in greater concentration, then the pH will be below the pK_a
• If the conjugate base is in greater concentration, then the pH will be higher than the pK_a
• In our example [A^-] > [HA] and pH > pK_a thus OK.

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In addition to these exercises you should work on the text materials.

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© R A Paselk

Last modified 9 December 2008