Question

Determine the pH during the titration of 34.6 mL of 0.350 M dimethylamine ((CH₃)₂NH , K_b = 5.9×10^-4) by 0.350M HCl at the following points.

(a) Before the addition of any HCl-?

(b) After the addition of 14.0 mL of HCl-?

(c) At the titration midpoint -?

(d) At the equivalence point -?

(e) After adding 50.5 mL of HCl-?

Answer 1

(a) The

pH generated by the dissociation of DMA in water:

DMA + H2O = DMA + + OH-

Kb = x ^ 2 / 0.35 - x = 5.9x10 ^ 4

Clearing we have:

x ^ 2 + 5.9x10 ^ -4 * x-2.07x10 ^ -4 = 0

Applying equation of the second degree we have:

x = [OH-] = 0.014

pOH = - Log (0.014) = 1.85

pH = 14 - 1.85 = 12.15

(b) We calculate the moles of added HCl and the moles of DMA:

DMA + HCl = DMA + + Cl-

n HCl = 0.35 * 0.014 = 0.0049 moles = n DMA +

n DMA = 0.35 * 0.0346 = 0.01211 moles

n DMA final = 0.01211 - 0.0049 = 0.00712 moles

With a total volume of 48.6 mL, we calculated the concentrations:

[DMA] = 0.00712 / 0.0486 = 0.148

[DMA +] = 0.0049 / 0.0486 = 0.1

We calculate the pKa:

Ka * Kb = Kw

Ka = Kw / Kb = 10 ^ -14 / 5.9x10 ^ -4 = 1.69 x10 ^ -11

pKa = - Log (1.69 x10 ^ -11) = 10.77

According to the Henderson-Hasselblach equation, we have:

pH = pKa + Log [DMA] / [DMA +] = 10.77 + Log (0.148 / 0.1) = 10.44

(c) At the midpoint of equivalence:

pH = pKa = 10.77

(d) At the equivalence point, the total volume is 69.2 mL, the moles of HCl added are 0.01211 moles, the concentration of DMA + produced is calculated:

[DMA +] = 0.01211 / 0.0692 = 0.175

The DMA + reacts with the water forming DMA and H3O + so

DMA + + H2O = DMA + H3O +

Ka = x ^ 2 / 0.175-x = 1.69x10 ^ -11

Clearing:

x ^ 2 + 1.69x10 ^ -4 * x - 2.96x10 ^ 5 = 0

Applying equation of the second degree:

X = [H3O +] = 5.36x10 ^ -3

pH = -Log (5.36x10 ^ -3) = 2.27

(e) In this step, the total volume is 85.1 mL, the initial moles of DMA are 0.01211, we calculate the moles of HCl added:

n Initial HCl = 0.35 * 0.0505 = 0.0177 moles

n Final HCl = 0.0177 - 0.01211 = 0.00559 moles

Calculation of concentration:

[HCl] = [H3O +] = 0.00559 / 0.0851 = 0.066 M

pH = - Log (0.066) = 1.18.