12. An unknown concentration solution (50.00 mL) of the salt, trimethylammonium
ID: 550705 • Letter: 1
Question
12. An unknown concentration solution (50.00 mL) of the salt, trimethylammonium chloride, is titrated with 0.1014 M NaOH. It requires 37.44 mL to reach the end point. Ka for trimethylamine = 1.58x10-10 (a) (I pt) What was the concentration of the original trimethylammonium chloride solution? (b) (no points, but you need this info)What is the volume of the solution at the end point? (c(1 pt) What is the pH of the solution at the end point? (Hint: write a balanced chemical equation for the titration reaction and examine the products. The titration reaction proceeds > 99.99% to products)Explanation / Answer
(CH3)3N: + HCl -----> (CH3)3NHCl(salt)
(CH3)3NHCl + NaOH ----> (CH3)3N: + NaCl + H2O
a) no of mol of NaOH reacted = 0.1014*37.44 = 3.8 mmol
volume of (CH3)3NHCl taken = 50 ml
Molarity(M) = n/v = 3.8/50 = 0.076 M
b) volume at the end point = 50+37.44 = 87.44 ml
C) (CH3)3NHCl + NaOH ----> (CH3)3N: + NaCl + H2O
total salt converted in to trimethyl amine.
concentration of trimethyl amine = 3.8/87.44 = 0.0434 M
pH = 14 - 1/2(pkb-logC)
pkb of trimethyl amine = 14-(-log(1.58*10^-10)) = 4.2
pH = 14 - 1/2(4.2-log0.0434) = 11.22
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