Question

Consider the titration of 100.0 mL of 0.100 M HCN by 0.100 M KOH at 25°C.
K_a for HCN = 6.2×10^-10.

Part 1
Calculate the pH after 0.0 mL of KOH has been added.

pH =

Part 2
Calculate the pH after 50.0 mL of KOH has been added.

pH =

Part 3
Calculate the pH after 75.0 mL of KOH has been added.

pH =

Part 4
Calculate the pH at the equivalence point.

pH =

Part 5
Calculate the pH after 125 mL of KOH has been added.

pH =

Answer 1

1)when 0.0 mL of KOH is added

HCN dissociates as:

HCN -----> H+ + CN-

0.1 0 0

0.1-x x x

Ka = [H+][CN-]/[HCN]

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 ((6.2*10^-10)*0.1) = 7.874*10^-6

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

so, x = 7.874*10^-6 M

use:

pH = -log [H+]

= -log (7.874*10^-6)

= 5.1038

Answer: 5.10

2)when 50.0 mL of KOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 50 mL

mol(HCN) = M(HCN) * V(HCN)

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

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 50 mL = 5 mmol

We have:

mol(HCN) = 10 mmol

mol(KOH) = 5 mmol

5 mmol of both will react

excess HCN remaining = 5 mmol

Volume of Solution = 100 + 50 = 150 mL

[HCN] = 5 mmol/150 mL = 0.0333M

[CN-] = 5/150 = 0.0333M

They form acidic buffer

acid is HCN

conjugate base is CN-

Ka = 6.2*10^-10

pKa = - log (Ka)

= - log(6.2*10^-10)

= 9.208

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.208+ log {3.333*10^-2/3.333*10^-2}

= 9.208

Answer: 9.21

3)when 75.0 mL of KOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 75 mL

mol(HCN) = M(HCN) * V(HCN)

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

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 75 mL = 7.5 mmol

We have:

mol(HCN) = 10 mmol

mol(KOH) = 7.5 mmol

7.5 mmol of both will react

excess HCN remaining = 2.5 mmol

Volume of Solution = 100 + 75 = 175 mL

[HCN] = 2.5 mmol/175 mL = 0.0143M

[CN-] = 7.5/175 = 0.0429M

They form acidic buffer

acid is HCN

conjugate base is CN-

Ka = 6.2*10^-10

pKa = - log (Ka)

= - log(6.2*10^-10)

= 9.208

use:

pH = pKa + log {[conjugate base]/[acid]}

= 9.208+ log {4.286*10^-2/1.429*10^-2}

= 9.685

Answer: 9.68

4)

find the volume of KOH used to reach equivalence point

M(HCN)*V(HCN) =M(KOH)*V(KOH)

0.1 M *100.0 mL = 0.1M *V(KOH)

V(KOH) = 100 mL

Given:

M(HCN) = 0.1 M

V(HCN) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 100 mL

mol(HCN) = M(HCN) * V(HCN)

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

mol(KOH) = M(KOH) * V(KOH)

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

We have:

mol(HCN) = 10 mmol

mol(KOH) = 10 mmol

10 mmol of both will react to form CN- and H2O

CN- here is strong base

CN- formed = 10 mmol

Volume of Solution = 100 + 100 = 200 mL

Kb of CN- = Kw/Ka = 1*10^-14/6.2*10^-10 = 1.613*10^-5

concentration ofCN-,c = 10 mmol/200 mL = 0.05M

CN- dissociates as

CN- + H2O -----> HCN + OH-

0.05 0 0

0.05-x x x

Kb = [HCN][OH-]/[CN-]

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 ((1.613*10^-5)*5*10^-2) = 8.98*10^-4

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

1.613*10^-5 = x^2/(5*10^-2-x)

8.065*10^-7 - 1.613*10^-5 *x = x^2

x^2 + 1.613*10^-5 *x-8.065*10^-7 = 0

This is quadratic equation (ax^2+bx+c=0)

a = 1

b = 1.613*10^-5

c = -8.065*10^-7

Roots can be found by

x = {-b + sqrt(b^2-4*a*c)}/2a

x = {-b - sqrt(b^2-4*a*c)}/2a

b^2-4*a*c = 3.226*10^-6

roots are :

x = 8.9*10^-4 and x = -9.061*10^-4

since x can't be negative, the possible value of x is

x = 8.9*10^-4

[OH-] = x = 8.9*10^-4 M

use:

pOH = -log [OH-]

= -log (8.9*10^-4)

= 3.0506

use:

PH = 14 - pOH

= 14 - 3.0506

= 10.9494

Answer: 10.95

5)when 125.0 mL of KOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 100 mL

M(KOH) = 0.1 M

V(KOH) = 125 mL

mol(HCN) = M(HCN) * V(HCN)

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

mol(KOH) = M(KOH) * V(KOH)

mol(KOH) = 0.1 M * 125 mL = 12.5 mmol

We have:

mol(HCN) = 10 mmol

mol(KOH) = 12.5 mmol

10 mmol of both will react

excess KOH remaining = 2.5 mmol

Volume of Solution = 100 + 125 = 225 mL

[OH-] = 2.5 mmol/225 mL = 0.0111 M

use:

pOH = -log [OH-]

= -log (1.111*10^-2)

= 1.9542

use:

PH = 14 - pOH

= 14 - 1.9542

= 12.0458

Answer: 12.05