Question

1) Find the pH during the titration of 20.00 mL of 0.1000 M ammonia (NH3), after adding the following volumes of 0.1000 M HCI. (2.5 pts each). In order to get credits in each question you must show your work. NO WORK=NO GRADE. Your work must be readable and easy to follow. Don't forget the correct use of units. a) Om b) 10.00 mL c) 15.00 ml d) 19.00 mL a) 19.95 ml f)20.00 mL B) 20.05 m. h) 25.00 mL

help calculating this please!

Answer 1

1)

a)when 0.0 mL of HCl is added

NH3 dissociates as:

NH3 +H2O -----> NH4+ + OH-

0.1 0 0

0.1-x x x

Kb = [NH4+][OH-]/[NH3]

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 ((1.8*10^-6)*0.1) = 4.243*10^-4

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

so, x = 4.243*10^-4 M

So, [OH-] = x = 4.243*10^-4 M

use:

pOH = -log [OH-]

= -log (4.243*10^-4)

= 3.3724

use:

PH = 14 - pOH

= 14 - 3.3724

= 10.6276

Answer: 10.63

b)when 10.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 10 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 10 mL = 1 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 1 mmol

mol(NH3) = 2 mmol

1 mmol of both will react

excess NH3 remaining = 1 mmol

Volume of Solution = 10 + 20 = 30 mL

[NH3] = 1 mmol/30 mL = 0.0333 M

[NH4+] = 1 mmol/30 mL = 0.0333 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-6

pKb = - log (Kb)

= - log(1.8*10^-6)

= 5.745

use:

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

= 5.745+ log {3.333*10^-2/3.333*10^-2}

= 5.745

use:

PH = 14 - pOH

= 14 - 5.7447

= 8.2553

Answer: 8.26

c)when 15.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 15 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 15 mL = 1.5 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 1.5 mmol

mol(NH3) = 2 mmol

1.5 mmol of both will react

excess NH3 remaining = 0.5 mmol

Volume of Solution = 15 + 20 = 35 mL

[NH3] = 0.5 mmol/35 mL = 0.0143 M

[NH4+] = 1.5 mmol/35 mL = 0.0429 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-6

pKb = - log (Kb)

= - log(1.8*10^-6)

= 5.745

use:

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

= 5.745+ log {4.286*10^-2/1.429*10^-2}

= 6.222

use:

PH = 14 - pOH

= 14 - 6.2218

= 7.7782

Answer: 7.78

d)when 19.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 19 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 19 mL = 1.9 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 1.9 mmol

mol(NH3) = 2 mmol

1.9 mmol of both will react

excess NH3 remaining = 0.1 mmol

Volume of Solution = 19 + 20 = 39 mL

[NH3] = 0.1 mmol/39 mL = 0.0026 M

[NH4+] = 1.9 mmol/39 mL = 0.0487 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-6

pKb = - log (Kb)

= - log(1.8*10^-6)

= 5.745

use:

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

= 5.745+ log {4.872*10^-2/2.564*10^-3}

= 7.023

use:

PH = 14 - pOH

= 14 - 7.0235

= 6.9765

Answer: 6.98

e)when 19.95 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 19.95 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 19.95 mL = 1.995 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 1.995 mmol

mol(NH3) = 2 mmol

1.995 mmol of both will react

excess NH3 remaining = 0.005 mmol

Volume of Solution = 19.95 + 20 = 39.95 mL

[NH3] = 0.005 mmol/39.95 mL = 0.0001 M

[NH4+] = 1.995 mmol/39.95 mL = 0.0499 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-6

pKb = - log (Kb)

= - log(1.8*10^-6)

= 5.745

use:

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

= 5.745+ log {4.994*10^-2/1.252*10^-4}

= 8.346

use:

PH = 14 - pOH

= 14 - 8.3457

= 5.6543

Answer: 5.65

f)when 20.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 20 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 20 mL = 2 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 2 mmol

mol(NH3) = 2 mmol

2 mmol of both will react to form NH4+ and H2O

NH4+ here is strong acid

NH4+ formed = 2 mmol

Volume of Solution = 20 + 20 = 40 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-6 = 5.556*10^-9

concentration ofNH4+,c = 2 mmol/40 mL = 0.05 M

NH4+ + H2O -----> NH3 + H+

5*10^-2 0 0

5*10^-2-x x x

Ka = [H+][NH3]/[NH4+]

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 ((5.556*10^-9)*5*10^-2) = 1.667*10^-5

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

so, x = 1.667*10^-5 M

[H+] = x = 1.667*10^-5 M

use:

pH = -log [H+]

= -log (1.667*10^-5)

= 4.7782

Answer: 4.78

g)when 20.05 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 20.05 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 20.05 mL = 2.005 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 2.005 mmol

mol(NH3) = 2 mmol

2 mmol of both will react

excess HCl remaining = 0.005 mmol

Volume of Solution = 20.05 + 20 = 40.05 mL

[H+] = 0.005 mmol/40.05 mL = 0.0001 M

use:

pH = -log [H+]

= -log (1.248*10^-4)

= 3.9036

Answer: 3.90

h)when 25.0 mL of HCl is added

Given:

M(HCl) = 0.1 M

V(HCl) = 25 mL

M(NH3) = 0.1 M

V(NH3) = 20 mL

mol(HCl) = M(HCl) * V(HCl)

mol(HCl) = 0.1 M * 25 mL = 2.5 mmol

mol(NH3) = M(NH3) * V(NH3)

mol(NH3) = 0.1 M * 20 mL = 2 mmol

We have:

mol(HCl) = 2.5 mmol

mol(NH3) = 2 mmol

2 mmol of both will react

excess HCl remaining = 0.5 mmol

Volume of Solution = 25 + 20 = 45 mL

[H+] = 0.5 mmol/45 mL = 0.0111 M

use:

pH = -log [H+]

= -log (1.111*10^-2)

= 1.9542

Answer: 1.95

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