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mylen [45]
3 years ago
11

What is the pH at the equivalence point in the titration of a 25.7 mL sample of a 0.370 M aqueous nitrous acid solution with a 0

.491 M aqueous potassium hydroxide solution? pH =
Chemistry
1 answer:
expeople1 [14]3 years ago
3 0

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 = \frac{[OH^-][HNO_2]}{[NO_2]}

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 = \frac{[X][X]}{[0.211 -X]}

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>

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Answer:

Kc = 1.54e - 31 / 2.61e - 24

Explanation:

1 )   N_{2}(gas) + O_{2}(gas)\rightarrow 2NO(gas)  ; Kc = 1.54e - 31

2)   N_{2}(gas) + 1/2O_{2}(gas)\rightarrow N_{2}O(gas)  ; Kc = 2.16e - 24

   upon reversing  ( 2 )  equation

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    now adding 1 and reversed equation (2)

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                  N_{2}O(gas) + 1/2O_{2}(gas)\rightarrow 2NO(gas)  Kc = 1.54e-31 × 1/2.61e - 24

       equilibrium constant of equation (3) is -

            Kc = 1.54e - 31 / 2.61e - 24

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