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Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka =...

Question

Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka =...

Determine the pH during the titration of 21.6 mL of 0.406 M hydrocyanic acid (Ka = 4.0×10-10) by 0.429 M NaOH at the following points.

(a) Before the addition of any NaOH _______

(b) After the addition of 5.30 mL of NaOH _________

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

(d) At the equivalence point ________

(e) After the addition of 30.7 mL of NaOH ________

Answers

a)when 0.0 mL of NaOH is added

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)when 5.3 mL of NaOH is added

Given:

M(HCN) = 0.406 M

V(HCN) = 21.6 mL

M(NaOH) = 0.429 M

V(NaOH) = 5.3 mL

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

mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol

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

mol(NaOH) = 0.429 M * 5.3 mL = 2.2737 mmol

We have:

mol(HCN) = 8.7696 mmol

mol(NaOH) = 2.2737 mmol

2.2737 mmol of both will react

excess HCN remaining = 6.4959 mmol

Volume of Solution = 21.6 + 5.3 = 26.9 mL

[HCN] = 6.4959 mmol/26.9 mL = 0.2415M

[CN-] = 2.2737/26.9 = 0.0845M

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 {8.452*10^-2/0.2415}

= 8.942

Answer: 8.94

c)

At half equivalence point:

pH = pKa

= 9.398

Answer: 9.40

d)

find the volume of NaOH used to reach equivalence point

M(HCN)*V(HCN) =M(NaOH)*V(NaOH)

0.406 M *21.6 mL = 0.429M *V(NaOH)

V(NaOH) = 20.442 mL

Given:

M(HCN) = 0.406 M

V(HCN) = 21.6 mL

M(NaOH) = 0.429 M

V(NaOH) = 20.442 mL

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

mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol

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

mol(NaOH) = 0.429 M * 20.442 mL = 8.7696 mmol

We have:

mol(HCN) = 8.7696 mmol

mol(NaOH) = 8.7696 mmol

8.7696 mmol of both will react to form CN- and H2O

CN- here is strong base

CN- formed = 8.7696 mmol

Volume of Solution = 21.6 + 20.442 = 42.042 mL

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

concentration ofCN-,c = 8.7696 mmol/42.042 mL = 0.2086M

CN- dissociates as

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

0.2086 0 0

0.2086-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)*0.2086) = 2.284*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/(0.2086-x)

5.215*10^-6 - 2.5*10^-5 *x = x^2

x^2 + 2.5*10^-5 *x-5.215*10^-6 = 0

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

a = 1

b = 2.5*10^-5

c = -5.215*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 = 2.086*10^-5

roots are :

x = 2.271*10^-3 and x = -2.296*10^-3

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

x = 2.271*10^-3

[OH-] = x = 2.271*10^-3 M

use:

pOH = -log [OH-]

= -log (2.271*10^-3)

= 2.6438

use:

PH = 14 - pOH

= 14 - 2.6438

= 11.3562

Answer: 11.36

e)when 30.7 mL of NaOH is added

Given:

M(HCN) = 0.406 M

V(HCN) = 21.6 mL

M(NaOH) = 0.429 M

V(NaOH) = 30.7 mL

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

mol(HCN) = 0.406 M * 21.6 mL = 8.7696 mmol

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

mol(NaOH) = 0.429 M * 30.7 mL = 13.1703 mmol

We have:

mol(HCN) = 8.7696 mmol

mol(NaOH) = 13.1703 mmol

8.7696 mmol of both will react

excess NaOH remaining = 4.4007 mmol

Volume of Solution = 21.6 + 30.7 = 52.3 mL

[OH-] = 4.4007 mmol/52.3 mL = 0.0841 M

use:

pOH = -log [OH-]

= -log (8.414*10^-2)

= 1.075

use:

PH = 14 - pOH

= 14 - 1.075

= 12.925

Answer: 12.93


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