How would the calculated concentration of the hci be affected if the sodium hydroxide were poured from ma beaker that contained
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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 = <em>45.1mL</em>
Potassium nitrite is in equilibrium with water, thus:
NO₂⁻ + H₂O ⇄ HNO₂ + OH⁻
Where equilibrium constant, Kb, is defined as:
Kb = 1.41x10⁻¹¹ =
In equilibrium, molarity of each compound are:
[NO₂⁻]: 0.00951mol/0.00451L - X = 0.211M - X
[HNO₂]: X
[OH⁻]: X
<em>Where X is reaction coordinate</em>
Replacing in Kb:
1.41x10⁻¹¹ =
0 = X² + 1.41x10⁻¹¹X - 2.97x10⁻¹²
Solving for X:
X = -1.72x10⁻⁶ <em>FALSE ANSWER. There is no negative concentrations.</em>
X = 1.72x10⁻⁶. <em>Right answer.</em>
That means:
[OH⁻]: 1.72x10⁻⁶M
As pOH is -log [OH⁻] and pH = 14-pOH:
pOH = 5.76; <em>pH = 8.24</em>