Question

Calculate the pH for the titration of 100 ml of 0.100M cocaine (Kb=2.6x10^-6) with 0.200M HNO3 at the folllowing points : a) 0ml, b) 25ml, c) 50ml and d) 60ml

Answer 1

1)when 0.0 mL of HNO3 is added

B dissociates as:

B +H2O -----> BH+ + OH-

0.1 0 0

0.1-x x x

Kb = [BH+][OH-]/[B]

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 ((2.6*10^-6)*0.1) = 5.099*10^-4

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

so, x = 5.099*10^-4 M

So, [OH-] = x = 5.099*10^-4 M

use:

pOH = -log [OH-]

= -log (5.099*10^-4)

= 3.2925

use:

PH = 14 - pOH

= 14 - 3.2925

= 10.7075

2)when 25.0 mL of HNO3 is added

Given:

M(HNO3) = 0.2 M

V(HNO3) = 25 mL

M(B) = 0.1 M

V(B) = 100 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 25 mL = 5 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HNO3) = 5 mmol

mol(B) = 10 mmol

5 mmol of both will react

excess B remaining = 5 mmol

Volume of Solution = 25 + 100 = 125 mL

[B] = 5 mmol/125 mL = 0.04 M

[BH+] = 5 mmol/125 mL = 0.04 M

They form basic buffer

base is B

conjugate acid is BH+

Kb = 2.6*10^-6

pKb = - log (Kb)

= - log(2.6*10^-6)

= 5.585

use:

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

= 5.585+ log {4*10^-2/4*10^-2}

= 5.585

use:

PH = 14 - pOH

= 14 - 5.585

= 8.415

3)when 50.0 mL of HNO3 is added

Given:

M(HNO3) = 0.2 M

V(HNO3) = 50 mL

M(B) = 0.1 M

V(B) = 100 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 50 mL = 10 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HNO3) = 10 mmol

mol(B) = 10 mmol

10 mmol of both will react to form BH+ and H2O

BH+ here is strong acid

BH+ formed = 10 mmol

Volume of Solution = 50 + 100 = 150 mL

Ka of BH+ = Kw/Kb = 1.0E-14/2.6E-6 = 3.846*10^-9

concentration ofBH+,c = 10 mmol/150 mL = 0.0667 M

BH+ + H2O -----> B + H+

6.667*10^-2 0 0

6.667*10^-2-x x x

Ka = [H+][B]/[BH+]

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 ((3.846*10^-9)*6.667*10^-2) = 1.601*10^-5

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

so, x = 1.601*10^-5 M

[H+] = x = 1.601*10^-5 M

use:

pH = -log [H+]

= -log (1.601*10^-5)

= 4.7955

4)when 60.0 mL of HNO3 is added

Given:

M(HNO3) = 0.2 M

V(HNO3) = 60 mL

M(B) = 0.1 M

V(B) = 100 mL

mol(HNO3) = M(HNO3) * V(HNO3)

mol(HNO3) = 0.2 M * 60 mL = 12 mmol

mol(B) = M(B) * V(B)

mol(B) = 0.1 M * 100 mL = 10 mmol

We have:

mol(HNO3) = 12 mmol

mol(B) = 10 mmol

10 mmol of both will react

excess HNO3 remaining = 2 mmol

Volume of Solution = 60 + 100 = 160 mL

[H+] = 2 mmol/160 mL = 0.0125 M

use:

pH = -log [H+]

= -log (1.25*10^-2)

= 1.9031