Question

50.0 mL of 0.100-M hydrogen cyanide (K_a = 6.20×10^-10) is titrated with 0.100-M NaOH.

What is the initial pH of the hydrogen cyanide solution?

What is the pH of the solution after 10.0 mL NaOH has been added?

What is the pH of the solution after a total of 25.0 mL NaOH has been added?

What is the pH of the solution after a total of 40.0 mL NaOH has been added?

What is the pH of the solution after a total of 50.0 mL NaOH has been added?

What is the pH of the solution after a total of 60.0 mL NaOH has been added?

Answer 1

1)when 0.0 mL of NaOH 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 10.0 mL of NaOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 10 mL

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

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

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 10 mL = 1 mmol

We have:

mol(HCN) = 5 mmol

mol(NaOH) = 1 mmol

1 mmol of both will react

excess HCN remaining = 4 mmol

Volume of Solution = 50 + 10 = 60 mL

[HCN] = 4 mmol/60 mL = 0.0667M

[CN-] = 1/60 = 0.0167M

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 {1.667*10^-2/6.667*10^-2}

= 8.606

Answer: 8.61

3)when 25.0 mL of NaOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 25 mL

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

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

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 25 mL = 2.5 mmol

We have:

mol(HCN) = 5 mmol

mol(NaOH) = 2.5 mmol

2.5 mmol of both will react

excess HCN remaining = 2.5 mmol

Volume of Solution = 50 + 25 = 75 mL

[HCN] = 2.5 mmol/75 mL = 0.0333M

[CN-] = 2.5/75 = 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

4)when 40.0 mL of NaOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 40 mL

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

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

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 40 mL = 4 mmol

We have:

mol(HCN) = 5 mmol

mol(NaOH) = 4 mmol

4 mmol of both will react

excess HCN remaining = 1 mmol

Volume of Solution = 50 + 40 = 90 mL

[HCN] = 1 mmol/90 mL = 0.0111M

[CN-] = 4/90 = 0.0444M

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 {4.444*10^-2/1.111*10^-2}

= 9.81

Answer: 9.81

5)when 50.0 mL of NaOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 50 mL

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

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

mol(NaOH) = M(NaOH) * V(NaOH)

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

We have:

mol(HCN) = 5 mmol

mol(NaOH) = 5 mmol

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

CN- here is strong base

CN- formed = 5 mmol

Volume of Solution = 50 + 50 = 100 mL

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

concentration ofCN-,c = 5 mmol/100 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

6)when 60.0 mL of NaOH is added

Given:

M(HCN) = 0.1 M

V(HCN) = 50 mL

M(NaOH) = 0.1 M

V(NaOH) = 60 mL

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

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

mol(NaOH) = M(NaOH) * V(NaOH)

mol(NaOH) = 0.1 M * 60 mL = 6 mmol

We have:

mol(HCN) = 5 mmol

mol(NaOH) = 6 mmol

5 mmol of both will react

excess NaOH remaining = 1 mmol

Volume of Solution = 50 + 60 = 110 mL

[OH-] = 1 mmol/110 mL = 0.0091 M

use:

pOH = -log [OH-]

= -log (9.091*10^-3)

= 2.0414

use:

PH = 14 - pOH

= 14 - 2.0414

= 11.9586

Answer: 11.96