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

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Debora [2.8K] · Answer 1 · 2023-09-04T17:51:01+03:00

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⁺ = $\frac{0.00572}{0.0446}$ ⇒ 0.128 M,

(C₂H₅)₃NH⁺ + H₃O⁺ ⇄ (C₂H₅)₃N + H₃O⁺

let's assume, (C₂H₅)₃N = $x$ ,

⇒ H₃O⁺ = $x$ , and

⇒ (C₂H₅)₃NH⁺ = 0.128 - $x$

∴ Now, k = kw / kb

⇒ k = 1.9 × 10⁻¹¹

and, k = [(C₂H₅)₃N][H₃O⁺] / (C₂H₅)₃NH⁺

⇒ 1.9 × 10⁻¹¹ = $x^{2}$ / (0.128 - $x$ )

Thus, $x$ = 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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