Answer:
pH = 8.24
Explanation:
Nitrous acid (HNO₂) reacts with KOH, thus:
HNO₂ + KOH → KNO₂ + H₂O
Moles of HNO₂ are:
0.0257mL ₓ (0.370mol / L) = 0.00951moles.
In equivalence point, the complete moles of nitrous acid reacts with KOH producing potassium nitrite. There are needed:
0.00951mol ₓ (1L / 0.491mol) = 0.01937L ≡ 19.4mL of 0.491M KOH to reach equivalence point.
Total volume in equivalence point is: 19.4mL + 25.7mL = 45.1mL
Potassium nitrite is in equilibrium with water, thus:
NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻
Where equilibrium constant, Kb, is defined as:
Kb = 1.41x10⁻¹¹ = [tex]\frac{[OH^-][HNO_2]}{[NO_2]}[/tex]
In equilibrium, molarity of each compound are:
[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X
[HNO₂]: X
[OH⁻]: X
Where X is reaction coordinate
Replacing in Kb:
1.41x10⁻¹¹ = [tex]\frac{[X][X]}{[0.211 -X]}[/tex]
0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²
Solving for X:
X = -1.72x10⁻⁶ FALSE ANSWER. There is no negative concentrations.
X = 1.72x10⁻⁶. Right answer.
That means:
[OH⁻]: 1.72x10⁻⁶M
As pOH is -log [OH⁻] and pH = 14-pOH:
pOH = 5.76; pH = 8.24
