Question

A 50.0 mL sample of 0.10 M ethanolamine, C2H4(OH)NH2, is titrated with 0.2 M HI. Calculate the pH to one decimal place when the following volumes of titrant have been added. A 50.0 mL sample of 0.10 M ethanolamine, C2H4(OH)NH2, is titrated with 0.2 M HI. Calculate the pH to one decimal place when the following volumes of titrant have been added 0.00mL 19.0mL 25.0mL 34.0ml

SHOW WORK PLEASE

Answer 1

Ethanolamine = C2H4(OH)NH2 = 50.0ml of 0.10M

Kb of Ethanolamine = 3.2 x 10^-5

a) 0.00ml of addition

for weak bases

[OH-] = square root of KbXC

[OH-] = square root of (3.2x10^-5 x 0.10)

[OH-] = 1.789 x10^-3

-loh[OH-] = -log(1.789 x10^-3)

POH = 2.75

PH + POH = 14

PH = 14 - POH

PH = 14 -2.75

PH = 11.25

b) addtion of 19.0 ml of HI

HI= 19.0 mL of 0.2M

number of moles of HI = 0.2M x 0.019L= 0.0038 moles

Ethanolamine = C2H4(OH)NH2 = 50.0ml of 0.10M

number of moles of Ethanolamine = C2H4(OH)NH2 = 0.10M x 0.050L = 0.0050 moles

Kb = 3.2x10^-5

-log(Kb) = -log(3.2x10^-5)

PKb = 4.49

                                          C2H4(OH)NH2    + HI   --------------------------- C2H4(OH)NH3+I-

Initial                                 0.0050                 0.0038                                            0

change                            - 0.0038               - 0.0038                                     +0.0038

equilibrium                    0.0012                         0                                            +0.0038

POH = PKb + log[salt]/[base]

POH = 4.49 + log(0.0038/0.0012)

POH = 4.99

PH = 14 - 4.99 = 9.01

PH = 9.01

C) addition of 25 mL of HI

HI = 25 mL of 0.2M

number of moles of HI= 0.2M x 0.025L = 0.0050 moles

number of moles of Ethanolamine = 0.0050 moles

It is equivalent point because number of moles of base is equal to acid

at equivalent point

PH = 7 - 1/2 [PKb +logC]

total volume= 50.0mL+ 25mL = 75mL = 0.075L

C= number of moles/volume = 0.0050/0.075 = 0.0667 M

PH = 7 - 1/2 [ 4.49 + log(0.0667)]

PH = 5.34

d) additon of 34 mL of HI

HI = 34mL of 0.2M

number of moles of HI = 0.2M x 0.034L = 0.0068 moles

number of moles of base = 0.0050 moles

number of moles of acid is more than that of base

so the nature of the solution is acidic.

remaining number of moles of acid = 0.0068 - 0.0050 = 0.0018 moles

total volume = 50.0+34.0 = 84.0 mL = 0.084 L

[H+] = 0.0018/0.084 = 0.0214

[H+] = 0.0214M

-log[H+] = -log(0.0214)

PH = 1.67