Example:
What is the pH of a 1.0 M solution of acetic acid?
Solution:
CH3CO2H
CH3CO– + H+ so, 
| CH3CO2H | CH3CO– | H+ | |
| Initial | 1.0 | 0 | 0 |
| Change | –x | +x | +x |
| Equilibrium | 1.0 – x | +x | +x |

| x = 0.0042 = [H+] | (1.0 – x ≈ 1.0 if x is small) |
pH = –log [0.0042] = 
Diprotic and Triprotic Acids
Substances containing more than one acidic proton are called polyprotic acids. Diprotic acids contain two acidic protons, and triprotic acids contain three acidic protons. Acid protons dissociate one at a time and have different Ka and pKa constants. Carbonic acid (H2CO3) is a diprotic acid.

pH of Weak Bases
Problems involving weak bases are treated similarly to the problems with weak acids. Weak bases (B) accept a proton from water (H2O) to produce a conjugate acid (HB+) and hydroxide ions:
B + H2O
+ HB+ + OH– 
Buffers
A buffer is a solution of a weak base and its conjugate acid (also weak) that prevents drastic changes in pH. The weak base reacts with any H+ ions that could increase acidity, and the weak conjugate acid reacts with OH– ions that may increase the basicity of the solution.
Carbonic Acid/Bicarbonate Buffer: H2CO3/HCO3–
Blood pH must be maintained at a pH of 7.40 by a buffer system consisting of the couple H2CO3/HCO3–:
H2CO3
HCO3– + H+
Neutralization of acid: HCO3– + H+ → H2CO3
Neutralization of base: H2CO3 + NaOH → NaHCO3 + H2O
Phosphate Buffer: H2PO4–/HPO4–2
The principal buffer system inside cells in blood consists of the couple H2PO4–/HPO4–2:

Neutralization of acid: HPO42– + H+ → H2PO4–
Neutralization of base:H2PO4– + OH – → HPO42– + H2O
The pH of a buffer solution can be calculated by the Henderson-Hasselbalch equation, which states that the pH of a buffer solution has a value close to the value of the weak acid (pKa):

Example:
What is the pH of a carbonate buffer solution with 0.65 M H2CO3 and 0.25 M HCO3–?
Solution:
H2CO3
HCO3– + H+
| acid | base |

Practice problems for these concepts can be found at - pH of Acids and Bases Practice Questions
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[H+] =
pH = –log[H+] = –log (3.33333 * 10–12) = 11.477 = 11
and pKa = –log [Ka]
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