Ethylamine, CH3CH2NH2, is an organic base with pKb = 3.367 at 298 K. In an exper

Question

Ethylamine, CH3CH2NH2, is an organic base with pKb = 3.367 at 298 K. In an experiment, a 40.0 mL sample of 0.105 mol L?1 CH3CH2NH2 (aq) is titrated with 0.150 mol L?1 HI(aq) solution at 298 K. (A) Write a balanced chemical equation for the neutralization reaction upon which this titration is based, and indicate clearly which atom is being protonated. Then, calculate the equilibrium constant for the neutralization reaction. (Hint: To calculate the equilibrium constant, you may find it helpful to represent the neutralization reaction as the sum of two separate reactions.) (B) Calculate the pH, [CH3CH2NH2], and [CH3CH2NH3+] at the following stages of the titration. i) before the addition of any HI solution. ii) after the addition of 20.0 mL of HI solution. iii) at the equivalence point. iv) after the addition of 60.0 mL of HI solution

Explanation / Answer

You want the equilibrium constant of the following reaction: CH3COOH(aq) + OH-(aq) -----> CH3COO-(aq) + H2O(l) This reaction is the sum of these two reactions: CH3COOH(aq) -----> H+(aq) + CH3COO-(aq) H+(aq) + OH-(aq) -----> H2O(l) Therefore, the equilibrium constant of your reaction can be determined by multiplying the equilibrium constants of these two reactions given. That of the first is 1.8e-5 as you stated. That of the second is 1e14 (since it is the reverse of water's dissociation equlibrium with Kw = 1e-14). Therefore, multiplying gives 1.8e9 as the answer, which is the last one.

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