Humboldt State University ® Department of Chemistry

Richard A. Paselk

	Introductory Chemistry	Fall 2005
Exercises: pH, Buffers etc.	© R. Paselk 2005
	Worked Examples
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Example 1 What is the pH of a solution of 0.033 M H₂SO₄?

Strong acid for first dissociation only, so [H⁺] = 0.033 M

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.

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

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

Barium hydroxide = Ba(OH)₂

Strong base, so [OH^-] = 2 (0.055) = 0.11 M

Recall that [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, 12 is the power of ten, only the figures to the right are significant.

 

 

 

 

 

 

 

 

 

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

Only the first dissociation will occur:

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 HOAc = 0.15 M   x   x

Example 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,

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

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

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

pK_a = 7.20

substituting

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

pH = 7.02

Example 5 Calculate the pH of a buffer made up by dissolving 0.0230 moles ascorbic acid (H₂C₆H₆O₆) and 0.0440 moles of sodium citrate (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

 

 

 

 

Example 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

Example 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