Question

Determine the pH during the titration of 31.1 mL of 0.259 M diethylamine ((C2H>NH, K,-6.9x10') by 0.259 M HI at the following points (Assume the titration is done at 25 °C) Note that state symbols are not shown for species in this problem (a) Before the addition of any HI (b) After the addition of 12.9 mL of HI (c) At the titration midpoint (d) At the equivalence point (e) After adding 48.5 mL of HI

Answer 1

Volume of (C₂H₅)₂NH = 31.1 mL

millimoles of (C₂H₅)₂NH = Molarity * volume = 0.259 M * 31.1 mL = 8.055 mmoles

K_b = 6.9 *10^-4

pKb = -log K_b = - log (6.9 *10^-4) = 3.161

(a) Before the addition of any HI

pOH = 1/2 (pK_b -log C) = 1/2 ( 3.161 - log (0.259)) = 1.287

pH + pOH = 14

pH = 14 - 1.287 = 12.71

(b) After the addition of 12.9 mL of HI

mmol of HI = Molarity * volume = 0.259 M * 12.9 mL = 3.34 mmol

Reaction is:

Reaction: (C₂H₅)₂NH +HI <=> (C₂H₅)₂NH₂⁺ + I^-

Initial (M) ...8.055........3.34

Change (M) -3.34 .........-3.34 ... . . +3.34 ... +3.34

Equili. (M) 4.71 ............0...............3.34............3.34

pOH = pKb + log [salt]/[base] = 3.16 + log (3.34/4.71) = 3.01

pH = 14 - pOH = 14 - 3.01 = 10.99

(c) At the titration midpoint

pOH = pKb = 3.16

pH = 14 - pOH = 14 - 3.16 = 10.84

(d) At equivalence Point,

Volume of HI required for equivalence = Volume of (C₂H₅)₂NH as both have same concentration = 31.1mL

So, total volume of solution = 31.1 + 31.1 = 62.2 mL

[salt] = 8.055 moles/ 62.2 mL = 0.1295 M

pH = 1/2( pKw - pKb - log c) = 1/2 (14 - 3.16 - log (0.1295)) = 5.86

(e) Now, we have excess of HI

Volume excess of HI = 17.4 mL

mmoles of HI in excess = 0.259 * 17.4 mL = 4.51 mmoles

Total volume of solution = 79.6 mL

[HI] excess = [H⁺] = 4.51 mmol / 79.6 mL = 0.0567

pH = -log [H⁺] = 1.25