Question

Consider the titration of 25.0 mL of 0.175 M cyanic acid, HCNO, with 0.250 M LiOH. The Ka of HCNO is 3.5 * 10^-4.

What is the volume of LIOH required to reach the equivalence point?
b. calculate the pH after the following volumes of LIOH have been added:
a. 0mL
b. 5.0 mL
c. 8.75 mL
d. 17.5 mL

Answer 1

a)

find the volume of LiOH used to reach equivalence point

M(HCNO)*V(HCNO) =M(LiOH)*V(LiOH)

0.175 M *25.0 mL = 0.25M *V(LiOH)

V(LiOH) = 17.5 mL

Answer: 17.5 mL

b)

1)when 0.0 mL of LiOH is added

HCNO dissociates as:

HCNO -----> H+ + CNO-

0.175 0 0

0.175-x x x

Ka = [H+][CNO-]/[HCNO]

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 ((3.5*10^-4)*0.175) = 7.826*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

3.5*10^-4 = x^2/(0.175-x)

6.125*10^-5 - 3.5*10^-4 *x = x^2

x^2 + 3.5*10^-4 *x-6.125*10^-5 = 0

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

a = 1

b = 3.5*10^-4

c = -6.125*10^-5

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 = 2.451*10^-4

roots are :

x = 7.653*10^-3 and x = -8.003*10^-3

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

x = 7.653*10^-3

use:

pH = -log [H+]

= -log (7.653*10^-3)

= 2.1162

2)when 5.0 mL of LiOH is added

Given:

M(HCNO) = 0.175 M

V(HCNO) = 25 mL

M(LiOH) = 0.25 M

V(LiOH) = 5 mL

mol(HCNO) = M(HCNO) * V(HCNO)

mol(HCNO) = 0.175 M * 25 mL = 4.375 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.25 M * 5 mL = 1.25 mmol

We have:

mol(HCNO) = 4.375 mmol

mol(LiOH) = 1.25 mmol

1.25 mmol of both will react

excess HCNO remaining = 3.125 mmol

Volume of Solution = 25 + 5 = 30 mL

[HCNO] = 3.125 mmol/30 mL = 0.1042M

[CNO-] = 1.25/30 = 0.0417M

They form acidic buffer

acid is HCNO

conjugate base is CNO-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

use:

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

= 3.456+ log {4.167*10^-2/0.1042}

= 3.058

3)when 8.75 mL of LiOH is added

Given:

M(HCNO) = 0.175 M

V(HCNO) = 25 mL

M(LiOH) = 0.25 M

V(LiOH) = 8.75 mL

mol(HCNO) = M(HCNO) * V(HCNO)

mol(HCNO) = 0.175 M * 25 mL = 4.375 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.25 M * 8.75 mL = 2.1875 mmol

We have:

mol(HCNO) = 4.375 mmol

mol(LiOH) = 2.1875 mmol

2.1875 mmol of both will react

excess HCNO remaining = 2.1875 mmol

Volume of Solution = 25 + 8.75 = 33.75 mL

[HCNO] = 2.1875 mmol/33.75 mL = 0.0648M

[CNO-] = 2.1875/33.75 = 0.0648M

They form acidic buffer

acid is HCNO

conjugate base is CNO-

Ka = 3.5*10^-4

pKa = - log (Ka)

= - log(3.5*10^-4)

= 3.456

use:

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

= 3.456+ log {6.481*10^-2/6.481*10^-2}

= 3.456

4)when 17.5 mL of LiOH is added

Given:

M(HCNO) = 0.175 M

V(HCNO) = 25 mL

M(LiOH) = 0.25 M

V(LiOH) = 17.5 mL

mol(HCNO) = M(HCNO) * V(HCNO)

mol(HCNO) = 0.175 M * 25 mL = 4.375 mmol

mol(LiOH) = M(LiOH) * V(LiOH)

mol(LiOH) = 0.25 M * 17.5 mL = 4.375 mmol

We have:

mol(HCNO) = 4.375 mmol

mol(LiOH) = 4.375 mmol

4.375 mmol of both will react to form CNO- and H2O

CNO- here is strong base

CNO- formed = 4.375 mmol

Volume of Solution = 25 + 17.5 = 42.5 mL

Kb of CNO- = Kw/Ka = 1*10^-14/3.5*10^-4 = 2.857*10^-11

concentration ofCNO-,c = 4.375 mmol/42.5 mL = 0.1029M

CNO- dissociates as

CNO- + H2O -----> HCNO + OH-

0.1029 0 0

0.1029-x x x

Kb = [HCNO][OH-]/[CNO-]

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 ((2.857*10^-11)*0.1029) = 1.715*10^-6

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

so, x = 1.715*10^-6 M

[OH-] = x = 1.715*10^-6 M

use:

pOH = -log [OH-]

= -log (1.715*10^-6)

= 5.7657

use:

PH = 14 - pOH

= 14 - 5.7657

= 8.2343