Question

calculate the pH for the titration 20 mL of 0.500 M HF with 0.500 M KOH at the following volumes
ka, HF is 6.8E-4 using the Rice table
a.0 mL
b.10 mL
c.19 mL
d. 20 mL
e. 21 mL
f. 30 mL

Answer 1

1)when 0.0 mL of KOH is added

HF dissociates as:

HF -----> H+ + F-

0.5 0 0

0.5-x x x

Ka = [H+][F-]/[HF]

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 ((6.8*10^-4)*0.5) = 1.844*10^-2

since x is comparable c, our assumption is not correct

we need to solve this using Quadratic equation

Ka = x*x/(c-x)

6.8*10^-4 = x^2/(0.5-x)

3.4*10^-4 - 6.8*10^-4 *x = x^2

x^2 + 6.8*10^-4 *x-3.4*10^-4 = 0

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

a = 1

b = 6.8*10^-4

c = -3.4*10^-4

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 = 1.36*10^-3

roots are :

x = 1.81*10^-2 and x = -1.878*10^-2

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

x = 1.81*10^-2

use:

pH = -log [H+]

= -log (1.81*10^-2)

= 1.7423

2)when 10.0 mL of KOH is added

Given:

M(HF) = 0.5 M

V(HF) = 20 mL

M(KOH) = 0.5 M

V(KOH) = 10 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.5 M * 20 mL = 10 mmol

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

mol(KOH) = 0.5 M * 10 mL = 5 mmol

We have:

mol(HF) = 10 mmol

mol(KOH) = 5 mmol

5 mmol of both will react

excess HF remaining = 5 mmol

Volume of Solution = 20 + 10 = 30 mL

[HF] = 5 mmol/30 mL = 0.1667M

[F-] = 5/30 = 0.1667M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

use:

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

= 3.167+ log {0.1667/0.1667}

= 3.167

3)when 19.0 mL of KOH is added

Given:

M(HF) = 0.5 M

V(HF) = 20 mL

M(KOH) = 0.5 M

V(KOH) = 19 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.5 M * 20 mL = 10 mmol

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

mol(KOH) = 0.5 M * 19 mL = 9.5 mmol

We have:

mol(HF) = 10 mmol

mol(KOH) = 9.5 mmol

9.5 mmol of both will react

excess HF remaining = 0.5 mmol

Volume of Solution = 20 + 19 = 39 mL

[HF] = 0.5 mmol/39 mL = 0.0128M

[F-] = 9.5/39 = 0.2436M

They form acidic buffer

acid is HF

conjugate base is F-

Ka = 6.8*10^-4

pKa = - log (Ka)

= - log(6.8*10^-4)

= 3.167

use:

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

= 3.167+ log {0.2436/1.282*10^-2}

= 4.446

4)when 20.0 mL of KOH is added

Given:

M(HF) = 0.5 M

V(HF) = 20 mL

M(KOH) = 0.5 M

V(KOH) = 20 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.5 M * 20 mL = 10 mmol

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

mol(KOH) = 0.5 M * 20 mL = 10 mmol

We have:

mol(HF) = 10 mmol

mol(KOH) = 10 mmol

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

F- here is strong base

F- formed = 10 mmol

Volume of Solution = 20 + 20 = 40 mL

Kb of F- = Kw/Ka = 1*10^-14/6.8*10^-4 = 1.471*10^-11

concentration ofF-,c = 10 mmol/40 mL = 0.25M

F- dissociates as

F- + H2O -----> HF + OH-

0.25 0 0

0.25-x x x

Kb = [HF][OH-]/[F-]

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 ((1.471*10^-11)*0.25) = 1.917*10^-6

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

so, x = 1.917*10^-6 M

[OH-] = x = 1.917*10^-6 M

use:

pOH = -log [OH-]

= -log (1.917*10^-6)

= 5.7173

use:

PH = 14 - pOH

= 14 - 5.7173

= 8.2827

5)when 21.0 mL of KOH is added

Given:

M(HF) = 0.5 M

V(HF) = 20 mL

M(KOH) = 0.5 M

V(KOH) = 21 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.5 M * 20 mL = 10 mmol

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

mol(KOH) = 0.5 M * 21 mL = 10.5 mmol

We have:

mol(HF) = 10 mmol

mol(KOH) = 10.5 mmol

10 mmol of both will react

excess KOH remaining = 0.5 mmol

Volume of Solution = 20 + 21 = 41 mL

[OH-] = 0.5 mmol/41 mL = 0.0122 M

use:

pOH = -log [OH-]

= -log (1.22*10^-2)

= 1.9138

use:

PH = 14 - pOH

= 14 - 1.9138

= 12.0862

6)when 30.0 mL of KOH is added

Given:

M(HF) = 0.5 M

V(HF) = 20 mL

M(KOH) = 0.5 M

V(KOH) = 30 mL

mol(HF) = M(HF) * V(HF)

mol(HF) = 0.5 M * 20 mL = 10 mmol

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

mol(KOH) = 0.5 M * 30 mL = 15 mmol

We have:

mol(HF) = 10 mmol

mol(KOH) = 15 mmol

10 mmol of both will react

excess KOH remaining = 5 mmol

Volume of Solution = 20 + 30 = 50 mL

[OH-] = 5 mmol/50 mL = 0.1 M

use:

pOH = -log [OH-]

= -log (0.1)

= 1

use:

PH = 14 - pOH

= 14 - 1

= 13