Question

AP CHEM question! Can you please follow these steps for each: 1.) net ionic equation, 2.) ice table, 3.) pH= pKa + log (CB/WA)

A 35.0 mL sample of 0.150 M acetic acid (HC2H3O2) is titrated with 0.150 M NaOH solution. Calculate the pH after the following volumes of base have been added:

A.) 0 mL
B.) 17.5 mL
C.) 34.5 mL
D.) 35.0 mL
E.) 35.5 mL
F.) 50.0 mL

Answer 1

1)when 0.0 mL of NaOH is added

HC2H3O2 dissociates as:

HC2H3O2 -----> H+ + C2H3O2-

0.15 0 0

0.15-x x x

Ka = [H+][C2H3O2-]/[HC2H3O2]

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 ((1.75*10^-5)*0.15) = 1.62*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)

1.75*10^-5 = x^2/(0.15-x)

2.625*10^-6 - 1.75*10^-5 *x = x^2

x^2 + 1.75*10^-5 *x-2.625*10^-6 = 0

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

a = 1

b = 1.75*10^-5

c = -2.625*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 = 1.05*10^-5

roots are :

x = 1.611*10^-3 and x = -1.629*10^-3

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

x = 1.611*10^-3

use:

pH = -log [H+]

= -log (1.611*10^-3)

= 2.7928

2)when 17.5 mL of NaOH is added

Given:

M(HC2H3O2) = 0.15 M

V(HC2H3O2) = 35 mL

M(NaOH) = 0.15 M

V(NaOH) = 17.5 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NaOH) = 0.15 M * 17.5 mL = 2.625 mmol

We have:

mol(HC2H3O2) = 5.25 mmol

mol(NaOH) = 2.625 mmol

2.625 mmol of both will react

excess HC2H3O2 remaining = 2.625 mmol

Volume of Solution = 35 + 17.5 = 52.5 mL

[HC2H3O2] = 2.625 mmol/52.5 mL = 0.05M

[C2H3O2-] = 2.625/52.5 = 0.05M

They form acidic buffer

acid is HC2H3O2

conjugate base is C2H3O2-

Ka = 1.75*10^-5

pKa = - log (Ka)

= - log(1.75*10^-5)

= 4.757

use:

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

= 4.757+ log {5*10^-2/5*10^-2}

= 4.757

3)when 34.5 mL of NaOH is added

Given:

M(HC2H3O2) = 0.15 M

V(HC2H3O2) = 35 mL

M(NaOH) = 0.15 M

V(NaOH) = 34.5 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NaOH) = 0.15 M * 34.5 mL = 5.175 mmol

We have:

mol(HC2H3O2) = 5.25 mmol

mol(NaOH) = 5.175 mmol

5.175 mmol of both will react

excess HC2H3O2 remaining = 0.075 mmol

Volume of Solution = 35 + 34.5 = 69.5 mL

[HC2H3O2] = 0.075 mmol/69.5 mL = 0.0011M

[C2H3O2-] = 5.175/69.5 = 0.0745M

They form acidic buffer

acid is HC2H3O2

conjugate base is C2H3O2-

Ka = 1.75*10^-5

pKa = - log (Ka)

= - log(1.75*10^-5)

= 4.757

use:

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

= 4.757+ log {7.446*10^-2/1.079*10^-3}

= 6.596

4)when 35.0 mL of NaOH is added

Given:

M(HC2H3O2) = 0.15 M

V(HC2H3O2) = 35 mL

M(NaOH) = 0.15 M

V(NaOH) = 35 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NaOH) = 0.15 M * 35 mL = 5.25 mmol

We have:

mol(HC2H3O2) = 5.25 mmol

mol(NaOH) = 5.25 mmol

5.25 mmol of both will react to form C2H3O2- and H2O

C2H3O2- here is strong base

C2H3O2- formed = 5.25 mmol

Volume of Solution = 35 + 35 = 70 mL

Kb of C2H3O2- = Kw/Ka = 1*10^-14/1.75*10^-5 = 5.714*10^-10

concentration ofC2H3O2-,c = 5.25 mmol/70 mL = 0.075M

C2H3O2- dissociates as

C2H3O2- + H2O -----> HC2H3O2 + OH-

0.075 0 0

0.075-x x x

Kb = [HC2H3O2][OH-]/[C2H3O2-]

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 ((5.714*10^-10)*7.5*10^-2) = 6.547*10^-6

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

so, x = 6.547*10^-6 M

[OH-] = x = 6.547*10^-6 M

use:

pOH = -log [OH-]

= -log (6.547*10^-6)

= 5.184

use:

PH = 14 - pOH

= 14 - 5.184

= 8.816

5)when 35.5 mL of NaOH is added

Given:

M(HC2H3O2) = 0.15 M

V(HC2H3O2) = 35 mL

M(NaOH) = 0.15 M

V(NaOH) = 35.5 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NaOH) = 0.15 M * 35.5 mL = 5.325 mmol

We have:

mol(HC2H3O2) = 5.25 mmol

mol(NaOH) = 5.325 mmol

5.25 mmol of both will react

excess NaOH remaining = 0.075 mmol

Volume of Solution = 35 + 35.5 = 70.5 mL

[OH-] = 0.075 mmol/70.5 mL = 0.0011 M

use:

pOH = -log [OH-]

= -log (1.064*10^-3)

= 2.9731

use:

PH = 14 - pOH

= 14 - 2.9731

= 11.0269

6)when 50.0 mL of NaOH is added

Given:

M(HC2H3O2) = 0.15 M

V(HC2H3O2) = 35 mL

M(NaOH) = 0.15 M

V(NaOH) = 50 mL

mol(HC2H3O2) = M(HC2H3O2) * V(HC2H3O2)

mol(HC2H3O2) = 0.15 M * 35 mL = 5.25 mmol

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

mol(NaOH) = 0.15 M * 50 mL = 7.5 mmol

We have:

mol(HC2H3O2) = 5.25 mmol

mol(NaOH) = 7.5 mmol

5.25 mmol of both will react

excess NaOH remaining = 2.25 mmol

Volume of Solution = 35 + 50 = 85 mL

[OH-] = 2.25 mmol/85 mL = 0.0265 M

use:

pOH = -log [OH-]

= -log (2.647*10^-2)

= 1.5772

use:

PH = 14 - pOH

= 14 - 1.5772

= 12.4228