Question

10 attempts left Check my work Be sure to answer all parts Report HI A 10.0-ml solution of 0.540 M NH, is titrated with a 0.180 M HCl solution. Calculate the pH after the following additions of the HCl solution: Guided (a) 0.00 mL (b) 10.0 mL (e) 30.0 mL @?@@@@ ONE (d) 40.0 mL

Answer 1

a)when 0.0 mL of HCl is added

NH3 dissociates as:

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

0.54 0 0

0.54-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)*0.54) = 3.118*10^-3

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

so, x = 3.118*10^-3 M

So, [OH-] = x = 3.118*10^-3 M

use:

pOH = -log [OH-]

= -log (3.118*10^-3)

= 2.5062

use:

PH = 14 - pOH

= 14 - 2.5062

= 11.4938

Answer: 11.49

b)when 10.0 mL of HCl is added

Given:

M(HCl) = 0.18 M

V(HCl) = 10 mL

M(NH3) = 0.54 M

V(NH3) = 10 mL

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

mol(HCl) = 0.18 M * 10 mL = 1.8 mmol

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

mol(NH3) = 0.54 M * 10 mL = 5.4 mmol

We have:

mol(HCl) = 1.8 mmol

mol(NH3) = 5.4 mmol

1.8 mmol of both will react

excess NH3 remaining = 3.6 mmol

Volume of Solution = 10 + 10 = 20 mL

[NH3] = 3.6 mmol/20 mL = 0.18 M

[NH4+] = 1.8 mmol/20 mL = 0.09 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 {9*10^-2/0.18}

= 4.444

use:

PH = 14 - pOH

= 14 - 4.4437

= 9.5563

Answer: 9.56

c)when 30.0 mL of HCl is added

Given:

M(HCl) = 0.18 M

V(HCl) = 30 mL

M(NH3) = 0.54 M

V(NH3) = 10 mL

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

mol(HCl) = 0.18 M * 30 mL = 5.4 mmol

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

mol(NH3) = 0.54 M * 10 mL = 5.4 mmol

We have:

mol(HCl) = 5.4 mmol

mol(NH3) = 5.4 mmol

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

NH4+ here is strong acid

NH4+ formed = 5.4 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 = 5.4 mmol/40 mL = 0.135 M

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

0.135 0 0

0.135-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)*0.135) = 8.66*10^-6

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

so, x = 8.66*10^-6 M

[H+] = x = 8.66*10^-6 M

use:

pH = -log [H+]

= -log (8.66*10^-6)

= 5.0625

Answer: 5.06

d)when 40.0 mL of HCl is added

Given:

M(HCl) = 0.18 M

V(HCl) = 40 mL

M(NH3) = 0.54 M

V(NH3) = 10 mL

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

mol(HCl) = 0.18 M * 40 mL = 7.2 mmol

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

mol(NH3) = 0.54 M * 10 mL = 5.4 mmol

We have:

mol(HCl) = 7.2 mmol

mol(NH3) = 5.4 mmol

5.4 mmol of both will react

excess HCl remaining = 1.8 mmol

Volume of Solution = 40 + 10 = 50 mL

[H+] = 1.8 mmol/50 mL = 0.036 M

use:

pH = -log [H+]

= -log (3.6*10^-2)

= 1.4437

Answer: 1.44