Question

titration of 50.0mL 0.50M NH3 with 0.50 M HCL. Kb(NH3)=1.8x10^-5
Plot the pH of the solution after the addition of the following volumes of HCL using ICE tables.
0.00mL, 10.00mL, 20.0mL, 25.0mL, 30.0mL, 40.0mL, 50.0mL,, 60.0mL

Answer 1

1)when 0.0 mL of HCl is added

NH3 dissociates as:

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

0.5 0 0

0.5-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^-5)*0.5) = 3*10^-3

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

so, x = 3*10^-3 M

So, [OH-] = x = 3*10^-3 M

use:

pOH = -log [OH-]

= -log (3*10^-3)

= 2.5229

use:

PH = 14 - pOH

= 14 - 2.5229

= 11.4771

2)when 10.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 10 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 10 mL = 5 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 5 mmol

mol(NH3) = 25 mmol

5 mmol of both will react

excess NH3 remaining = 20 mmol

Volume of Solution = 10 + 50 = 60 mL

[NH3] = 20 mmol/60 mL = 0.3333 M

[NH4+] = 5 mmol/60 mL = 0.0833 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {8.333*10^-2/0.3333}

= 4.143

use:

PH = 14 - pOH

= 14 - 4.1427

= 9.8573

3)when 20.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 20 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 20 mL = 10 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 10 mmol

mol(NH3) = 25 mmol

10 mmol of both will react

excess NH3 remaining = 15 mmol

Volume of Solution = 20 + 50 = 70 mL

[NH3] = 15 mmol/70 mL = 0.2143 M

[NH4+] = 10 mmol/70 mL = 0.1429 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.1429/0.2143}

= 4.569

use:

PH = 14 - pOH

= 14 - 4.5686

= 9.4314

4)when 25.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 25 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 25 mL = 12.5 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 12.5 mmol

mol(NH3) = 25 mmol

12.5 mmol of both will react

excess NH3 remaining = 12.5 mmol

Volume of Solution = 25 + 50 = 75 mL

[NH3] = 12.5 mmol/75 mL = 0.1667 M

[NH4+] = 12.5 mmol/75 mL = 0.1667 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.1667/0.1667}

= 4.745

use:

PH = 14 - pOH

= 14 - 4.7447

= 9.2553

5)when 30.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 30 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 30 mL = 15 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 15 mmol

mol(NH3) = 25 mmol

15 mmol of both will react

excess NH3 remaining = 10 mmol

Volume of Solution = 30 + 50 = 80 mL

[NH3] = 10 mmol/80 mL = 0.125 M

[NH4+] = 15 mmol/80 mL = 0.1875 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.1875/0.125}

= 4.921

use:

PH = 14 - pOH

= 14 - 4.9208

= 9.0792

6)when 40.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 40 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 40 mL = 20 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 20 mmol

mol(NH3) = 25 mmol

20 mmol of both will react

excess NH3 remaining = 5 mmol

Volume of Solution = 40 + 50 = 90 mL

[NH3] = 5 mmol/90 mL = 0.0556 M

[NH4+] = 20 mmol/90 mL = 0.2222 M

They form basic buffer

base is NH3

conjugate acid is NH4+

Kb = 1.8*10^-5

pKb = - log (Kb)

= - log(1.8*10^-5)

= 4.745

use:

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

= 4.745+ log {0.2222/5.556*10^-2}

= 5.347

use:

PH = 14 - pOH

= 14 - 5.3468

= 8.6532

7)when 50.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 50 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 50 mL = 25 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 25 mmol

mol(NH3) = 25 mmol

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

NH4+ here is strong acid

NH4+ formed = 25 mmol

Volume of Solution = 50 + 50 = 100 mL

Ka of NH4+ = Kw/Kb = 1.0E-14/1.8E-5 = 5.556*10^-10

concentration ofNH4+,c = 25 mmol/100 mL = 0.25 M

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

0.25 0 0

0.25-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^-10)*0.25) = 1.179*10^-5

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

so, x = 1.179*10^-5 M

[H+] = x = 1.179*10^-5 M

use:

pH = -log [H+]

= -log (1.179*10^-5)

= 4.9287

8)when 60.0 mL of HCl is added

Given:

M(HCl) = 0.5 M

V(HCl) = 60 mL

M(NH3) = 0.5 M

V(NH3) = 50 mL

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

mol(HCl) = 0.5 M * 60 mL = 30 mmol

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

mol(NH3) = 0.5 M * 50 mL = 25 mmol

We have:

mol(HCl) = 30 mmol

mol(NH3) = 25 mmol

25 mmol of both will react

excess HCl remaining = 5 mmol

Volume of Solution = 60 + 50 = 110 mL

[H+] = 5 mmol/110 mL = 0.0455 M

use:

pH = -log [H+]

= -log (4.545*10^-2)

= 1.3424