Understanding the Chemistry

* NaNO₂ is a salt: It dissociates completely in water to form Na⁺ and NO₂⁻ ions.

* NO₂⁻ is a weak base: It reacts with water to produce a small amount of OH⁻ ions, making the solution slightly basic.

Steps to Calculate pH

1. Write the equilibrium reaction for the weak base:

NO₂⁻(aq) + H₂O(l) ⇌ HNO₂(aq) + OH⁻(aq)

2. Set up an ICE table (Initial, Change, Equilibrium):

| | NO₂⁻ | HNO₂ | OH⁻ |

|-----------|-------------|-------------|-----------|

| Initial | 0.35 M | 0 M | 0 M |

| Change | -x | +x | +x |

| Equilibrium| 0.35 - x | x | x |

3. Write the Kb expression:

Kb = [HNO₂][OH⁻] / [NO₂⁻]

4. Substitute the equilibrium concentrations from the ICE table into the Kb expression:

2.5 x 10⁻¹¹ = (x)(x) / (0.35 - x)

5. Since Kb is very small, we can assume x is negligible compared to 0.35:

2.5 x 10⁻¹¹ ≈ (x)(x) / 0.35

6. Solve for x, which represents the [OH⁻]:

x² = 8.75 x 10⁻¹²

x = [OH⁻] ≈ 2.96 x 10⁻⁶ M

7. Calculate pOH:

pOH = -log[OH⁻] = -log(2.96 x 10⁻⁶) ≈ 5.53

8. Calculate pH using the relationship between pH and pOH:

pH + pOH = 14

pH = 14 - pOH = 14 - 5.53 ≈ 8.47

Therefore, the pH of a 0.35 M NaNO₂ solution is approximately 8.47.