Question

Determine the pH during the titration of 22.7 mL of 0.406 M hydrocyanic acid (K_a = 4.0×10^-10) by 0.492 M KOH at the following points.

(a) Before the addition of any KOH

(b) After the addition of 4.60 mL of KOH

(c) At the half-equivalence point (the titration midpoint)

(d) At the equivalence point

(e) After the addition of 28.1 mL of KOH

Answer 1

a)

HCN dissociates as:

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

0.406 0 0

0.406-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 ((4*10^-10)*0.406) = 1.274*10^-5

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

so, x = 1.274*10^-5 M

use:

pH = -log [H+]

= -log (1.274*10^-5)

= 4.8947

Answer: 4.89

b)

Given:

M(HCN) = 0.406 M

V(HCN) = 22.7 mL

M(KOH) = 0.492 M

V(KOH) = 4.6 mL

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

mol(HCN) = 0.406 M * 22.7 mL = 9.2162 mmol

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

mol(KOH) = 0.492 M * 4.6 mL = 2.2632 mmol

We have:

mol(HCN) = 9.2162 mmol

mol(KOH) = 2.2632 mmol

2.2632 mmol of both will react

excess HCN remaining = 6.953 mmol

Volume of Solution = 22.7 + 4.6 = 27.3 mL

[HCN] = 6.953 mmol/27.3 mL = 0.2547M

[CN-] = 2.2632/27.3 = 0.0829M

They form acidic buffer

acid is HCN

conjugate base is CN-

Ka = 4*10^-10

pKa = - log (Ka)

= - log(4*10^-10)

= 9.398

use:

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

= 9.398+ log {8.29*10^-2/0.2547}

= 8.91

Answer: 8.91

c)

At half equivalence point.

pH = pKa

use:

pKa = -log Ka

= -log (4*10^-10)

= 9.3979

Answer: 9.40

d)

find the volume of KOH used to reach equivalence point

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

0.406 M *22.7 mL = 0.492M *V(KOH)

V(KOH) = 18.7321 mL

Given:

M(HCN) = 0.406 M

V(HCN) = 22.7 mL

M(KOH) = 0.492 M

V(KOH) = 18.7321 mL

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

mol(HCN) = 0.406 M * 22.7 mL = 9.2162 mmol

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

mol(KOH) = 0.492 M * 18.7321 mL = 9.2162 mmol

We have:

mol(HCN) = 9.2162 mmol

mol(KOH) = 9.2162 mmol

9.2162 mmol of both will react to form CN- and H2O

CN- here is strong base

CN- formed = 9.2162 mmol

Volume of Solution = 22.7 + 18.7321 = 41.4321 mL

Kb of CN- = Kw/Ka = 1*10^-14/4*10^-10 = 2.5*10^-5

concentration ofCN-,c = 9.2162 mmol/41.4321 mL = 0.2224M

CN- dissociates as

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

0.2224 0 0

0.2224-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 ((2.5*10^-5)*0.2224) = 2.358*10^-3

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Kb = x*x/(c-x)

2.5*10^-5 = x^2/(0.2224-x)

5.561*10^-6 - 2.5*10^-5 *x = x^2

x^2 + 2.5*10^-5 *x-5.561*10^-6 = 0

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

a = 1

b = 2.5*10^-5

c = -5.561*10^-6

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 = 2.224*10^-5

roots are :

x = 2.346*10^-3 and x = -2.371*10^-3

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

x = 2.346*10^-3

[OH-] = x = 2.346*10^-3 M

use:

pOH = -log [OH-]

= -log (2.346*10^-3)

= 2.6297

use:

PH = 14 - pOH

= 14 - 2.6297

= 11.3703

Answer: 11.37

e)when 28.1 mL of KOH is added

Given:

M(HCN) = 0.406 M

V(HCN) = 22.7 mL

M(KOH) = 0.492 M

V(KOH) = 28.1 mL

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

mol(HCN) = 0.406 M * 22.7 mL = 9.2162 mmol

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

mol(KOH) = 0.492 M * 28.1 mL = 13.8252 mmol

We have:

mol(HCN) = 9.2162 mmol

mol(KOH) = 13.8252 mmol

9.2162 mmol of both will react

excess KOH remaining = 4.609 mmol

Volume of Solution = 22.7 + 28.1 = 50.8 mL

[OH-] = 4.609 mmol/50.8 mL = 0.0907 M

use:

pOH = -log [OH-]

= -log (9.073*10^-2)

= 1.0423

use:

PH = 14 - pOH

= 14 - 1.0423

= 12.9577

Answer: 12.96