Calculating pH of Titration (CH3)3N with HClO4
November 5th, 2008
Free chemistry Tutor Homework Help
Consider the titration of 20.0 mL of 0.0800 M (CH3)3N (a weak base; Kb = 6.40e-05) with 0.100 M HClO4.? Calculate the pH after the following volumes of titrant have been added:
(a) 0.0 mL
(b) 4.0 mL
(c) 8.0 mL
Answer
(a). Since no titrant has been added to the analyt so we only have 20.0 mL of 0.0800 M (CH3)3N thus the pH of this solution is calculated using this equation:
[OH-] = (Kb x M)1/2
[OH-] = (6.4.10-5x 0.0800)1/2
[OH-] = 2.26×10-2
pOH = - log 2.26×10-2 = 2.64
pH = 14- 2.64 = 11.36
b. 20.0 mL of 0.0800 M (CH3)3N is equal to 1.6 mmol of (CH3)3N and we add 4.0 mL HClO4 0.100 M that equal to 0.4 mmol. Thus 1.6 mmol (CH3)3N to be react with 0.4 mmol HClO4 this will be produce 0.4 mmol of (CH3)NHCLO4. The solution contains base and its salt, so this will be form a buffer solution
The balance reaction after with mole after reaction occur:
(CH3)3N + HClO4 -> (CH3)3NHClO4
1,2 mmol—0 mmol———0.4 mmol
We can use the henderson-Hasselbalch equation to calculate pH of the buffer solution
pH = pKa + log (base/salt)
Ka for (CH3)3N = Kw/Kb = 1×10-14/6.4.10-5 = 1.56.10-10
pKa = - log Ka = - log(1.56.10-10) = 9.8
pH = 9.8 + log (1.2 mmol/0.4 mmol) = 10
c. Using the same way that we use to solve the b problem we get:
(CH3)3N + HClO4 -> (CH3)3NHClO4
0.8 mmol—-0 mmol—-0.8 mmol
pH = 9.8 + log (0.8 mmol/0.8 mmol) = 9.8
Entry Filed under: Titration
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