Question

You are titrating 100.00 mL of 0.4200 M propanoic acid ( CH3CH2CO2H) with 1.000 M KOH Find the pH at the following points during the titration. The Ka for propanoic acid is 1.34 x 105 15. (32 pts) a) 0.00 mL of 1.000 M KOH b) 10.00 mL of 1.000 M KOH c) 21.00 mL of 1.000 M KOH d) 42.00 mL of 1.000 M KOH e) 60.00 mL of 1.000 M KOH f) Circle the best indicator for this acid-base titration Indicator In1 In2 In3 In4 Transition Range 1.00 2.00 2.00-3.00 3.00-4.00 4.00 5.00 Indicator In5 In6 In7 In8 Transition Range Indicator 5.00-6.00 6.00 7.00 7.00 8.00 8.00 9.00 In9 In10 In11 In12 Transition Range 9.00 10.00 10.00- 11.00 11.00 -12.00 12.00 13.00

Answer 1

1)when 0.0 mL of KOH is added

CH3CH2COOH dissociates as:

CH3CH2COOH -----> H+ + CH3CH2COO-

0.42 0 0

0.42-x x x

Ka = [H+][CH3CH2COO-]/[CH3CH2COOH]

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 ((4.5*10^-4)*0.42) = 1.375*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)

4.5*10^-4 = x^2/(0.42-x)

1.89*10^-4 - 4.5*10^-4 *x = x^2

x^2 + 4.5*10^-4 *x-1.89*10^-4 = 0

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

a = 1

b = 4.5*10^-4

c = -1.89*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 = 7.562*10^-4

roots are :

x = 1.352*10^-2 and x = -1.397*10^-2

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

x = 1.352*10^-2

use:

pH = -log [H+]

= -log (1.352*10^-2)

= 1.8689

2)when 10.0 mL of KOH is added

Given:

M(CH3CH2COOH) = 0.42 M

V(CH3CH2COOH) = 100 mL

M(KOH) = 1 M

V(KOH) = 10 mL

mol(CH3CH2COOH) = M(CH3CH2COOH) * V(CH3CH2COOH)

mol(CH3CH2COOH) = 0.42 M * 100 mL = 42 mmol

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

mol(KOH) = 1 M * 10 mL = 10 mmol

We have:

mol(CH3CH2COOH) = 42 mmol

mol(KOH) = 10 mmol

10 mmol of both will react

excess CH3CH2COOH remaining = 32 mmol

Volume of Solution = 100 + 10 = 110 mL

[CH3CH2COOH] = 32 mmol/110 mL = 0.2909M

[CH3CH2COO-] = 10/110 = 0.0909M

They form acidic buffer

acid is CH3CH2COOH

conjugate base is CH3CH2COO-

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.347

use:

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

= 3.347+ log {9.091*10^-2/0.2909}

= 2.842

3)when 21.0 mL of KOH is added

Given:

M(CH3CH2COOH) = 0.42 M

V(CH3CH2COOH) = 100 mL

M(KOH) = 1 M

V(KOH) = 21 mL

mol(CH3CH2COOH) = M(CH3CH2COOH) * V(CH3CH2COOH)

mol(CH3CH2COOH) = 0.42 M * 100 mL = 42 mmol

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

mol(KOH) = 1 M * 21 mL = 21 mmol

We have:

mol(CH3CH2COOH) = 42 mmol

mol(KOH) = 21 mmol

21 mmol of both will react

excess CH3CH2COOH remaining = 21 mmol

Volume of Solution = 100 + 21 = 121 mL

[CH3CH2COOH] = 21 mmol/121 mL = 0.1736M

[CH3CH2COO-] = 21/121 = 0.1736M

They form acidic buffer

acid is CH3CH2COOH

conjugate base is CH3CH2COO-

Ka = 4.5*10^-4

pKa = - log (Ka)

= - log(4.5*10^-4)

= 3.347

use:

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

= 3.347+ log {0.1736/0.1736}

= 3.347

4)when 42.0 mL of KOH is added

Given:

M(CH3CH2COOH) = 0.42 M

V(CH3CH2COOH) = 100 mL

M(KOH) = 1 M

V(KOH) = 42 mL

mol(CH3CH2COOH) = M(CH3CH2COOH) * V(CH3CH2COOH)

mol(CH3CH2COOH) = 0.42 M * 100 mL = 42 mmol

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

mol(KOH) = 1 M * 42 mL = 42 mmol

We have:

mol(CH3CH2COOH) = 42 mmol

mol(KOH) = 42 mmol

42 mmol of both will react to form CH3CH2COO- and H2O

CH3CH2COO- here is strong base

CH3CH2COO- formed = 42 mmol

Volume of Solution = 100 + 42 = 142 mL

Kb of CH3CH2COO- = Kw/Ka = 1*10^-14/4.5*10^-4 = 2.222*10^-11

concentration ofCH3CH2COO-,c = 42 mmol/142 mL = 0.2958M

CH3CH2COO- dissociates as

CH3CH2COO- + H2O -----> CH3CH2COOH + OH-

0.2958 0 0

0.2958-x x x

Kb = [CH3CH2COOH][OH-]/[CH3CH2COO-]

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 ((2.222*10^-11)*0.2958) = 2.564*10^-6

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

so, x = 2.564*10^-6 M

[OH-] = x = 2.564*10^-6 M

use:

pOH = -log [OH-]

= -log (2.564*10^-6)

= 5.5911

use:

PH = 14 - pOH

= 14 - 5.5911

= 8.4089

5)when 60.0 mL of KOH is added

Given:

M(CH3CH2COOH) = 0.42 M

V(CH3CH2COOH) = 100 mL

M(KOH) = 1 M

V(KOH) = 60 mL

mol(CH3CH2COOH) = M(CH3CH2COOH) * V(CH3CH2COOH)

mol(CH3CH2COOH) = 0.42 M * 100 mL = 42 mmol

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

mol(KOH) = 1 M * 60 mL = 60 mmol

We have:

mol(CH3CH2COOH) = 42 mmol

mol(KOH) = 60 mmol

42 mmol of both will react

excess KOH remaining = 18 mmol

Volume of Solution = 100 + 60 = 160 mL

[OH-] = 18 mmol/160 mL = 0.1125 M

use:

pOH = -log [OH-]

= -log (0.1125)

= 0.9488

use:

PH = 14 - pOH

= 14 - 0.9488

= 13.0512

only 4 parts at a time