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A 29.4 ml sample of 0.347 m triethylamine, (c2h5)3n, is titrated with 0.375 m hydrobromic acid. at the equivalence point, the pH is 5.81.
Given,
Volume of Triethylamine (C₂H₅)₃N = 29.4 ml = 0.0294 L
Molarity of (C₂H₅)₃N = 0.275 M
Molarity of HBr = 0.375M
Solution:
(C₂H₅)₃N + HBr ⇔ (C₂H₅)₃ NH⁺ + Br⁻
⇒ Volume of HBr = mole of (C₂H₅)₃N / Molarity of HBr
⇒ Volume of HBr = 0.00572 mol / 0.375 mol/l
⇒ Volume of HBr = 0.0152 L
∴ Total Volume = 0.0294 + 0.0152
⇒ Total Volume = 0.0446 L
Concentration of (C₂H₅)₃NH⁺ = ⇒ 0.128 M,
(C₂H₅)₃NH⁺ + H₃O⁺ ⇄ (C₂H₅)₃N + H₃O⁺
let's assume, (C₂H₅)₃N = ,
⇒ H₃O⁺ = , and
⇒ (C₂H₅)₃NH⁺ = 0.128 -
∴ Now, k = kw / kb
⇒ k = 1.9 × 10⁻¹¹
and, k = [(C₂H₅)₃N][H₃O⁺] / (C₂H₅)₃NH⁺
⇒ 1.9 × 10⁻¹¹ = / (0.128 -
)
Thus, = 1.55 × 10⁻⁶ M,
Hence, pH = - log[x]
⇒ - log [1.55 × 10⁻⁶ ]
⇒ - log (1.55) - log (10⁻⁶)
⇒ - log (1.55) + 6log10
⇒ - 0.19 + 6
⇒ 5.81 = pH
Therefore, pH = 5.81.
To learn more about equivalence point here
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