Question

27. 0.1 points 0/4 Submissions Used Consider the titration of 20.0 mL of 0.0100 MHON (a weak base: K - 1.1008) with 0.100 . Calculate the ph after the following volumes of the bes t (a) 0.0 ml (b) 4.0 mL (d) 12.0 ML (0) 16.0 ml PH-

Answer 1

a)when 0.0 mL of HI is added

HONH2 dissociates as:

HONH2 +H2O -----> HONH3+ + OH-

8*10^-2 0 0

8*10^-2-x x x

Kb = [HONH3+][OH-]/[HONH2]

Kb = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Kb = x*x/(c)

so, x = sqrt (Kb*c)

x = sqrt ((1.1*10^-8)*8*10^-2) = 2.966*10^-5

since c is much greater than x, our assumption is correct

so, x = 2.966*10^-5 M

So, [OH-] = x = 2.966*10^-5 M

use:

pOH = -log [OH-]

= -log (2.966*10^-5)

= 4.5278

use:

PH = 14 - pOH

= 14 - 4.5278

= 9.4722

Answer: 9.47

b)when 4.0 mL of HI is added

Given:

M(HI) = 0.1 M

V(HI) = 4 mL

M(HONH2) = 0.08 M

V(HONH2) = 20 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.1 M * 4 mL = 0.4 mmol

mol(HONH2) = M(HONH2) * V(HONH2)

mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HI) = 0.4 mmol

mol(HONH2) = 1.6 mmol

0.4 mmol of both will react

excess HONH2 remaining = 1.2 mmol

Volume of Solution = 4 + 20 = 24 mL

[HONH2] = 1.2 mmol/24 mL = 0.05 M

[HONH3+] = 0.4 mmol/24 mL = 0.0167 M

They form basic buffer

base is HONH2

conjugate acid is HONH3+

Kb = 1.1*10^-8

pKb = - log (Kb)

= - log(1.1*10^-8)

= 7.959

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 7.959+ log {1.667*10^-2/5*10^-2}

= 7.481

use:

PH = 14 - pOH

= 14 - 7.4815

= 6.5185

Answer: 6.52

c)when 8.0 mL of HI is added

Given:

M(HI) = 0.1 M

V(HI) = 8 mL

M(HONH2) = 0.08 M

V(HONH2) = 20 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.1 M * 8 mL = 0.8 mmol

mol(HONH2) = M(HONH2) * V(HONH2)

mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HI) = 0.8 mmol

mol(HONH2) = 1.6 mmol

0.8 mmol of both will react

excess HONH2 remaining = 0.8 mmol

Volume of Solution = 8 + 20 = 28 mL

[HONH2] = 0.8 mmol/28 mL = 0.0286 M

[HONH3+] = 0.8 mmol/28 mL = 0.0286 M

They form basic buffer

base is HONH2

conjugate acid is HONH3+

Kb = 1.1*10^-8

pKb = - log (Kb)

= - log(1.1*10^-8)

= 7.959

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 7.959+ log {2.857*10^-2/2.857*10^-2}

= 7.959

use:

PH = 14 - pOH

= 14 - 7.9586

= 6.0414

Answer: 6.04

d)when 12.0 mL of HI is added

Given:

M(HI) = 0.1 M

V(HI) = 12 mL

M(HONH2) = 0.08 M

V(HONH2) = 20 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.1 M * 12 mL = 1.2 mmol

mol(HONH2) = M(HONH2) * V(HONH2)

mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HI) = 1.2 mmol

mol(HONH2) = 1.6 mmol

1.2 mmol of both will react

excess HONH2 remaining = 0.4 mmol

Volume of Solution = 12 + 20 = 32 mL

[HONH2] = 0.4 mmol/32 mL = 0.0125 M

[HONH3+] = 1.2 mmol/32 mL = 0.0375 M

They form basic buffer

base is HONH2

conjugate acid is HONH3+

Kb = 1.1*10^-8

pKb = - log (Kb)

= - log(1.1*10^-8)

= 7.959

use:

pOH = pKb + log {[conjugate acid]/[base]}

= 7.959+ log {3.75*10^-2/1.25*10^-2}

= 8.436

use:

PH = 14 - pOH

= 14 - 8.4357

= 5.5643

Answer: 5.56

e)when 16.0 mL of HI is added

Given:

M(HI) = 0.1 M

V(HI) = 16 mL

M(HONH2) = 0.08 M

V(HONH2) = 20 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.1 M * 16 mL = 1.6 mmol

mol(HONH2) = M(HONH2) * V(HONH2)

mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HI) = 1.6 mmol

mol(HONH2) = 1.6 mmol

1.6 mmol of both will react to form HONH3+ and H2O

HONH3+ here is strong acid

HONH3+ formed = 1.6 mmol

Volume of Solution = 16 + 20 = 36 mL

Ka of HONH3+ = Kw/Kb = 1.0E-14/1.1E-8 = 9.091*10^-7

concentration ofHONH3+,c = 1.6 mmol/36 mL = 0.0444 M

HONH3+ + H2O -----> HONH2 + H+

4.444*10^-2 0 0

4.444*10^-2-x x x

Ka = [H+][HONH2]/[HONH3+]

Ka = x*x/(c-x)

Assuming x can be ignored as compared to c

So, above expression becomes

Ka = x*x/(c)

so, x = sqrt (Ka*c)

x = sqrt ((9.091*10^-7)*4.444*10^-2) = 2.01*10^-4

since c is much greater than x, our assumption is correct

so, x = 2.01*10^-4 M

[H+] = x = 2.01*10^-4 M

use:

pH = -log [H+]

= -log (2.01*10^-4)

= 3.6968

Answer: 3.70

f)when 27.2 mL of HI is added

Given:

M(HI) = 0.1 M

V(HI) = 27.2 mL

M(HONH2) = 0.08 M

V(HONH2) = 20 mL

mol(HI) = M(HI) * V(HI)

mol(HI) = 0.1 M * 27.2 mL = 2.72 mmol

mol(HONH2) = M(HONH2) * V(HONH2)

mol(HONH2) = 0.08 M * 20 mL = 1.6 mmol

We have:

mol(HI) = 2.72 mmol

mol(HONH2) = 1.6 mmol

1.6 mmol of both will react

excess HI remaining = 1.12 mmol

Volume of Solution = 27.2 + 20 = 47.2 mL

[H+] = 1.12 mmol/47.2 mL = 0.0237 M

use:

pH = -log [H+]

= -log (2.373*10^-2)

= 1.6247

Answer: 1.62