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Resuelva los siguientes problemas 1. Tiene que mostrar todos sus cálculos. calcule la concentración del ion de hidronio y el ph en acido nitroso que tiene una concentración de 3.1 x 10-3m. hno2 + H2O → H3O+ + NO2 ka = 7.1 x 10-4
Accepted Answer
The concentration of hydronium ion ([H3O+]) and pH of a 3.1 x 10^-3 M nitrous acid (HNO2) solution can be calculated as follows:
1. Set up an ICE table:
| | HNO2 | H3O+ | NO2- |
|-----------|--------|--------|--------|
| Initial | 3.1E-3 | 0 | 0 |
| Change | -x | +x | +x |
| Equilibrium| 3.1E-3-x | x | x |
2. Write the Ka expression:
Ka = [H3O+][NO2-] / [HNO2]
3. Substitute the equilibrium concentrations into the Ka expression:
7.1 x 10^-4 = (x)(x) / (3.1 x 10^-3 - x)
4. Since Ka is small, we can assume that x is negligible compared to 3.1 x 10^-3:
7.1 x 10^-4 = (x^2) / (3.1 x 10^-3)
5. Solve for x:
x^2 = (7.1 x 10^-4)(3.1 x 10^-3)
x = √(7.1 x 10^-4)(3.1 x 10^-3) ≈ 1.48 x 10^-3
Therefore, the concentration of hydronium ion [H3O+] is approximately 1.48 x 10^-3 M.
6. Calculate the pH:
pH = -log[H3O+]
pH = -log(1.48 x 10^-3) ≈ 2.83
Therefore, the concentration of hydronium ion is approximately 1.48 x 10^-3 M and the pH is approximately 2.83.
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