Answer:
a. 0.180M of C₆H₅NH₂
b. 0.0887M C₆H₅NH₃⁺
c. pH = 2.83
Explanation:
a. Based in the chemical equation:
C₆H₅NH₂(aq) + HCl(aq) → C₆H₅NH₃⁺(aq) + Cl⁻(aq)
<em>1 mole of aniline reacts per mole of HCl</em>
Moles required to reach equivalence point are:
Moles HCl = 0.02567L ₓ (0.175mol / L) = 4.492x10⁻³ moles HCl = moles C₆H₅NH₂
As the original solution had a volume of 25.0mL = 0.0250L:
4.492x10⁻³ moles C₆H₅NH₂ / 0.0250L = 0.180M of C₆H₅NH₂
b. At equivalence point, moles of C₆H₅NH₃⁺ are equal to initial moles of C₆H₅NH₂, that is 4.492x10⁻³ moles
But now, volume is 25.0mL + 25.67mL = 50.67mL = 0.05067L. Thus, molar concentration of C₆H₅NH₃⁺ is:
[C₆H₅NH₃⁺] = 4.492x10⁻³ moles / 0.05067L = 0.0887M C₆H₅NH₃⁺
c. At equivalence point you have just 0.0887M C₆H₅NH₃⁺ in solution. C₆H₅NH₃⁺ has as equilibrium in water:
C₆H₅NH₃⁺(aq) + H₂O(l) → C₆H₅NH₂ + H₃O⁺
Where Ka = Kw / Kb = 1x10⁻¹⁴ / 4.0x10⁻¹⁰ =
<em>2.5x10⁻⁵ = [C₆H₅NH₂] [H₃O⁺] / [C₆H₅NH₃⁺]</em>
When the system reaches equilibrium, molar concentrations are:
[C₆H₅NH₃⁺] = 0.0887M - X
[C₆H₅NH₂] = X
[H₃O⁺] = X
Replacing in Ka formula:
2.5x10⁻⁵ = [X] [X] / [0.0887M - X]
2.2175x10⁻⁶ - 2.5x10⁻⁵X = X²
0 = X² + 2.5x10⁻⁵X - 2.2175x10⁻⁶
Solving for X:
X = -0.0015 → False solution. There is no negative concentrations.
X = 0.001477 → Right solution.
As [H₃O⁺] = X, [H₃O⁺] = 0.001477
Knowing pH = -log [H₃O⁺]
pH = -log 0.001477
<h3>pH = 2.83</h3>
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