Understanding the Chemistry

* NaCN is the salt of a weak acid (HCN) and a strong base (NaOH). This means the solution will be basic.

* The CN⁻ ion will hydrolyze in water, reacting with water to produce hydroxide ions (OH⁻) and make the solution basic.

The Steps

1. Write the hydrolysis reaction:

CN⁻(aq) + H₂O(l) ⇌ HCN(aq) + OH⁻(aq)

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

| | CN⁻ | HCN | OH⁻ |

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

| Initial | 0.028 M | 0 | 0 |

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

| Equilibrium | 0.028-x | x | x |

3. Write the Kb expression:

Kb = ([HCN][OH⁻]) / [CN⁻]

4. Calculate Kb:

Kw = Ka * Kb

Kb = Kw / Ka = (1.0 x 10⁻¹⁴) / (4.9 x 10⁻¹⁰) = 2.04 x 10⁻⁵

5. Substitute the equilibrium concentrations into the Kb expression:

2.04 x 10⁻⁵ = (x * x) / (0.028 - x)

6. Since Kb is small, we can assume x is negligible compared to 0.028:

2.04 x 10⁻⁵ ≈ (x²) / 0.028

7. Solve for x (which represents [OH⁻]):

x² ≈ 2.04 x 10⁻⁵ * 0.028

x ≈ √(2.04 x 10⁻⁵ * 0.028)

x ≈ 7.54 x 10⁻⁴ M

8. Calculate pOH:

pOH = -log[OH⁻] = -log(7.54 x 10⁻⁴) ≈ 3.12

9. Calculate pH:

pH + pOH = 14

pH = 14 - pOH ≈ 14 - 3.12 ≈ 10.88

Therefore, the pH of a 0.028 M NaCN solution is approximately 10.88.