Answer:
(a) 10.62
(b) 2.82
(c) 1.95
Explanation:
The neutralization reaction in this question is
(CH3CH2)3N + HCl ⇒ (CH3CH2)3NH⁺ + Cl⁻
The problem can be solved by calculating the number of moles of triethylamine after addition of the portions of HCl. Since it is a weak base if it is not consumed completely, that is in excess we will have a buffer of a waek base. If its consumed completely the pH will be determined by the strong acid HCl.
The pOH for a buffer of a weak base is gven by
pOH = pKb + log [(CH3CH2)3NH⁺] / [(CH3CH2)3N]
(a) 11 mL of 0.100 M HCl
mol HCl = 0.011 L x 0.100 mol/L = 0.0011 mol HCl
mol (CH3CH2)3N reacted = 0.0011 mol
mol (CH3CH2)3NH⁺ produced = 0.0011 mol
mol (CH3CH2)3N initially = 0.020 L x 0.1000 mol/L 0.0020 mol
mol (CH3CH2)3N left = 0.0020 mol - 0.0011 = 0.0009 mol
pKb = - log Kb = - log (5.2 x 10⁻⁴) = 3.284
Now we can compute pOH,
pOH = 3.284 + log ( 0.0011 / 0.0009 ) = 3.37
pH = 14 - pOH = 14 - 3.37 = 10.62
(b) 20.60 mL HCl
mol HCl = 0.0206 L x 0.100 mol/L = 0.00206
mol (CH3CH2)3N consumed = 0.0020 mol
This is so because the acid will consume completely the 0.0020 mol of the weak base we had originally present.
Now the problem circumscribes to that of calculating the pH of the unreacted HCl
Total Vol = 0.0206 L + 0.02 L = 0.0406 L
mol HCl = 0.0206 L x .100 = 0.00206 mol
mol HCl left = 0.00206 mol - 0.0020 mol = 0.00006 mol
[HCl] = 0.00006 mol / 0.0406 L = 0.0015 M
Since HCl is a strong acid ( 100 % ionization) :
pH = - log [H⁺] = - log ( 0.0015 ) = 2.82
(c) We will compute the pH in the same way we did for part (b)
mol HCl = 0.025 L x 0.100 mol/L = 0.0025 mol
mol HCl left = 0.0025 mol - 0.0020 mol = 0.0005
Total Volume = 0.020 L + 0.025 L = 0.045 L
[HCl] = 0.0005 mol / 0.045 L = 0.111
pH = - log ( 0.111) = 1.95
