1. Chapter 14 Acids and Bases
14.9 Buffers
Learning Goal Describe the role of buffers in maintaining the pH of a solution; calculate the pH of a buffer.
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2. When an acid or base is added to water, the pH changes drastically
Buffers
When an acid or base is added
to water, the pH changes drastically
to a buffer solution, the pH resists change
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3. a combination of the acid–base conjugate pair
Buffers
A buffer is
a combination of the acid–base conjugate pair
a weak acid and salt of its conjugate base
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4. HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
Buffer System
Buffer systems resist a change in pH by shifting the equilibrium in the other direction.
HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2−(aq)
When H3O+ is added, it reacts with C2H3O2−.
When OH− is added, it reacts with HC2H3O2.
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5. Which combination makes a buffer solution? A. HCl and KCl
Learning Check
Which combination makes a buffer solution?
A. HCl and KCl
B. H2CO3 and NaHCO3
C. H3PO4 and NaCl
D. CH3COOH and LiCH3COO
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6. Which combination makes a buffer solution? H2CO3 and NaHCO3
Buffer, weak acid, and salt of its conjugate base.
CH3COOH and LiCH3COO
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7. Calculating the pH of a Buffer
Given the Ka expression for a weak acid, HA and its conjugate base A− can be rearranged to solve for [H3O+]. ×
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8. Guide to Calculating pH of a Buffer
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9. H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Learning Check
The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 × 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4− and HPO42−?
H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
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10. Solution
The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 × 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4− and HPO42−? Step 1 State the given and needed quantities.
Given
Need
Know
0.20 M H2PO4−
0.20 M HPO42− pH
H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
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11. H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Solution
The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 × 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4− and HPO42−?
H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Step 2 Write the Ka expression and rearrange for [H3O+] .
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12. H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Solution
The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 × 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4− and HPO42−?
H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq)
Step 3 Substitute [HA] and [A−] into Ka expression.
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13. Solution
The weak acid H2PO4- in a buffer H2PO4−/HPO42− has Ka = 6.2 × 10−8. What is the pH of the buffer if it is 0.20 M in both H2PO4− and HPO42−? H2PO4(aq) + H2O(l) HPO42(aq) + H3O+(aq) Step 4 Use [H3O+] to calculate pH. pH = −log[6.2 × 10−8] = 6.2 × 10−8 M = 7.21
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14. Chemistry Link to Health, Blood Buffer
Carbonic acid and bicarbonate regulate blood pH.
CO2(g) + H2CO3(aq) H3O+(aq) + HCO3−(aq)
When the body absorbs H3O+, it reacts with HCO3−, and the equilibrium shifts toward the reactants.
When OH− from foods and cellular processes is absorbed, it reacts with H2CO3, and the equilibrium shifts toward the products.
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15. Concept Map
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