Question

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Be sure to answer all parts A 10.0−mL solution of 0.690 M NH3 is titrated with a 0.230 M HCl solution. Calculate the pH after the following additions of the HCl solution: (a) 0.00 mL (b) 10.0 mL (c) 30.0 mL (d) 40.0 mL

Answer 1

1)when 0.0 mL of HCl is added

NH3 dissociates as:

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

0.69 0 0

0.69-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)*0.69) = 3.524*10^-3

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

so, x = 3.524*10^-3 M

So, [OH-] = x = 3.524*10^-3 M

use:

pOH = -log [OH-]

= -log (3.524*10^-3)

= 2.4529

use:

PH = 14 - pOH

= 14 - 2.4529

= 11.5471

2)when 10.0 mL of HCl is added

Given:

M(HCl) = 0.23 M

V(HCl) = 10 mL

M(NH3) = 0.69 M

V(NH3) = 10 mL

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

mol(HCl) = 0.23 M * 10 mL = 2.3 mmol

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

mol(NH3) = 0.69 M * 10 mL = 6.9 mmol

We have:

mol(HCl) = 2.3 mmol

mol(NH3) = 6.9 mmol

2.3 mmol of both will react

excess NH3 remaining = 4.6 mmol

Volume of Solution = 10 + 10 = 20 mL

[NH3] = 4.6 mmol/20 mL = 0.23 M

[NH4+] = 2.3 mmol/20 mL = 0.115 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 {0.115/0.23}

= 4.444

use:

PH = 14 - pOH

= 14 - 4.4437

= 9.5563

3)when 30.0 mL of HCl is added

Given:

M(HCl) = 0.23 M

V(HCl) = 30 mL

M(NH3) = 0.69 M

V(NH3) = 10 mL

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

mol(HCl) = 0.23 M * 30 mL = 6.9 mmol

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

mol(NH3) = 0.69 M * 10 mL = 6.9 mmol

We have:

mol(HCl) = 6.9 mmol

mol(NH3) = 6.9 mmol

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

NH4+ here is strong acid

NH4+ formed = 6.9 mmol

Volume of Solution = 30 + 10 = 40 mL

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

concentration ofNH4+,c = 6.9 mmol/40 mL = 0.1725 M

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

0.1725 0 0

0.1725-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)*0.1725) = 9.789*10^-6

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

so, x = 9.789*10^-6 M

[H+] = x = 9.789*10^-6 M

use:

pH = -log [H+]

= -log (9.789*10^-6)

= 5.0092

4)when 40.0 mL of HCl is added

Given:

M(HCl) = 0.23 M

V(HCl) = 40 mL

M(NH3) = 0.69 M

V(NH3) = 10 mL

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

mol(HCl) = 0.23 M * 40 mL = 9.2 mmol

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

mol(NH3) = 0.69 M * 10 mL = 6.9 mmol

We have:

mol(HCl) = 9.2 mmol

mol(NH3) = 6.9 mmol

6.9 mmol of both will react

excess HCl remaining = 2.3 mmol

Volume of Solution = 40 + 10 = 50 mL

[H+] = 2.3 mmol/50 mL = 0.046 M

use:

pH = -log [H+]

= -log (4.6*10^-2)

= 1.3372