1. Write the balanced chemical equation for the dissociation of nitric acid in water:
HNO3 + H2O → H3O+ + NO3-
2. Set up an ICE table to represent the initial concentrations, changes in concentrations, and equilibrium concentrations of the species in the solution:
| | HNO3 | H3O+ | NO3- |
|---|---|---|---|
| Initial | 0.0001M | 0M | 0M |
| Change | -x | +x | +x |
| Equilibrium | (0.0001-x)M | xM | xM |
3. Substitute the equilibrium concentrations into the expression for the acid dissociation constant (Ka) for nitric acid:
Ka = [H3O+][NO3-]/[HNO3]
Ka = x^2/(0.0001-x)
4. Since Ka for nitric acid is very large (2.0 x 10^6), we can assume that x is much smaller than 0.0001 and simplify the expression:
Ka ≈ x^2/0.0001
x^2 ≈ 2.0 x 10^(-8)
x ≈ 1.4 x 10^(-4)
5. Calculate the pH of the solution using the formula:
pH = -log[H3O+]
pH = -log(1.4 x 10^(-4))
pH ≈ 3.85
Therefore, the pH of a 0.0001M water solution of nitric acid is approximately 3.85.
