Chemical reactivity - a) Dissociation reaction with water - Ethanamine is a weak base. Show its dissociation in water to form ethanammonium ion and hydroxide ions. b) Neutralisation. - Amines are weak bases as they can accept a proton. The reaction will form a salt and water. i) Reaction with inorganic acids - eg. HCl. Write an equation to show the neutralisation of methanamine with HCI. 024 Year 12 Chemistry BSSS ii) [Harder: Finish this the equation: CH$_3$NHCH$_3$(aq) + HNO$_3$(aq) →] Neutralisation with an organic acid - Amines can also react with organic acids like carboxylic acids in a neutralisation reaction. BUT - under special conditions the formation of an amide and water is produced. The reaction needs a mild dehydrating agent, called DCC, in neutral conditions. Show the equation of the reaction of ethanoic acid and methanamine. 1
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In this reaction, the methanamine (CH$_3$NH$_2$) acts as a base and reacts with HCl, which is an acid. The products of this reaction are a salt, methanammonium chloride (CH$_3$NH$_3$Cl), and water (H$_2$O). Show more…
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Ethanamine, CH3CH2NH2, is a weak base. pKa(CH3CH2NH3+) = 10.6 (a) (i) Write an equation to show the reaction of ethanamine CH3CH2NH2 with water. CH3CH2NH2 + H2O ⇌ CH3CH2NH3+ + OH- (ii) Explain, with reference to the equation and the species in solution, what is meant by the terms weak and base. In the equation, ethanamine (CH3CH2NH2) reacts with water to form ethanammonium ion (CH3CH2NH3+) and hydroxide ion (OH-). A weak base is a substance that partially dissociates in water, meaning it does not completely ionize. In this case, ethanamine only partially reacts with water to form the ethanammonium ion and hydroxide ion. (b) (i) Determine the pKb for ethanamine. pKb = 14 - pKa pKb = 14 - 10.6 pKb = 3.4 (ii) Calculate the pH of a 0.109 mol L-1 solution of ethanamine. To calculate the pH, we need to find the pOH first. pOH = -log10[OH-] pOH = -log10(0.109) pOH = 0.962 pH + pOH = 14 pH = 14 - pOH pH = 14 - 0.962 pH = 13.038
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