Help calculating theoretical pH

CH3COOH or Acetic acid

OAc- is the shortcut for CH3COO- or the Acetate ion, the conjugate base of Acetic acid



1) VHCl = 0 mL (before addn of HCl)



Change of Concentrations Due to Dilution (M1V1 = M2V2)

[OAc-] = 0.100 M
50 mL
[OAc-] = (0.100)(50)/95 = 0.0526 M

[HOAc] = 0.100 M
45 mL
[HOAc] = (0.100)(45)/95 = 0.0474 M

Vtotal = 95 mL




You took 30.00 mL of the buffer:

[OAc-] = 0.0526M
30 mL
[OAc-] =
(0.0526)(30.00)
=
0.0522 M

30.25 + 0

[HOAc] = 0.0474 M
30 mL
[HOAc] =
(0.0474)(30.00)
=
0.0470 M

30.25 + 0




Plug these values in the ICE Table:

HOAc + H2O < = > H3O+ + OAc-]
Initial 0.0470 M 10-7 0.0522 M
Change -x +x +x
Equilibrium 0.0470 - x x 0.0522 + x



Then into the equilibrium expression:

Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0522 + x)
[HOAc] 0.0470 - x

5% of 0.0474 = 2.37 x 10-3

assume x is less than this value

1.76 x 10-5 = x (0.0522 )
0.0470

Solving for x = [H3O+]

[H3O+] = 1.58 x 10-5

pH = - log [H3O+] = 4.80 M



2) Addn of 0.100 M HCl

Calculate these concentrations for each volume of acid added corresponding to the increments added in Table 12.2 in the manual:

[OAc-] = 0.0526M
30 mL
[OAc-] =
(0.0526)(30.00)

30.25 + VHCl added

[HOAc] = 0.0474 M
30 mL
[HOAc] =
(0.0474)(30.00)

30.25 + VHCl added


Remember that the volumes are additive (i.e., the volume of HCl added after the second addition of the acid is 0.50 + 1.00 = 1.50 mL, third time is 1.50 + 1.50 = 3.0 mL and so on )



You are adding an acid to the buffer, calculate [HCl] for each volume of acid added:

[HCl]
=
0.100 M
x
V HCl added

30.25 + VHCl added


You will be adding this value to [HOAc] and subtracting it from [OAc-] in the ICE table:



Example: After addition of 0.50 mL of HCl, [HCl] = 0.00163

HOAc + H2O < = > H3O+ + OAc-]
Initial 0. 0462 + 0.00163 10-7 0.0513 - 0.00163
Change -x +x +x
Equilibrium 0.0478 - x x 0.0497 + x



Then into the equilibrium expression:

Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0497 + x)
[HOAc] 0.0478 - x

5% of 0.0478 = 2.39 x 10-3

assume x is less than this value

1.76 x 10-5 = x (0.0497 )
0.0478

Solving for x = [H3O+]

[H3O+] = 1.69 x 10-5 < 2.39 x 10-3

pH = - log [H3O+] = 4.77 M

Notice that the pH did not change that much compared to the original pH of 4.80 unlike the degassed water where the pH dropped from 7.0 to 2.1 with the addition of 0.25 mL of HCl.

Also, notice that the addition of an acid decreases the pH.



Part C: Addition of NaOH to Degassed Water

1) Without the base, the pH should be close to 7.0, [OH-] = 10-7

1) Without the base, the pH should be close to 7.0, [OH-] = 10-7

2) Calculate [NaOH] for each volume corresponding to each increment of base added in Table 12.3 on p.74 of the manual:

[NaOH]
=
0.100 M
x
V NaOH added

V total


Note that Vtotal = Vwater + Vindicator + VHCl added = 30.00 + 0.25 + V NaOH added

3) Calculate [OH-] and the corresponding pH:

NaOH is a strong acid and therefore it dissociates completely in water and therefore:

[NaOH] = [OH-]

pOH = - log [OH-]

pH = 14 - pOH



Part D: Addition of NaOH to Acetate Buffer (Weak acid/Conjugate base Combination)



1) VNaOH = 0 mL (before addn of NaOH)



Change of Concentrations Due to Dilution (M1V1 = M2V2)

[OAc-] = 0.100 M
50 mL
[OAc-] = (0.100)(50)/95 = 0.0526 M

[HOAc] = 0.100 M
45 mL
[HOAc] = (0.100)(45)/95 = 0.0474 M

Vtotal = 95 mL




You took 30.00 mL of the buffer:

[OAc-] = 0.0526M
30 mL
[OAc-] =
(0.0526)(30.00)
=
0.0522 M

30.25 + 0

[HOAc] = 0.0474 M
30 mL
[HOAc] =
(0.0474)(30.00)
=
0.0470 M

30.25 + 0




Plug these values in the ICE Table:

HOAc + H2O < = > H3O+ + OAc-]
Initial 0.0470 M 10-7 0.0522 M
Change -x +x +x
Equilibrium 0.0470 - x x 0.0522 + x



Then into the equilibrium expression:

Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0.0522 + x)
[HOAc] 0.0470 - x

5% of 0.0474 = 2.37 x 10-3

assume x is less than this value

1.76 x 10-5 = x (0.0522 )
0.0470

Solving for x = [H3O+]

[H3O+] = 1.58 x 10-5

pH = - log [H3O+] = 4.80 M (Same as in Part B)



2) Addn of 0.100 M NaOH

Calculate these concentrations for each volume of base added corresponding to the increments added in Table 12.4 in the manual:

[OAc-] = 0.0526M
30 mL
[OAc-] =
(0.0526)(30.00)

30.25 + VNaOH added

[HOAc] = 0.0474 M
30 mL
[HOAc] =
(0.0474)(30.00)

30.25 + VNaOH added


Remember that the volumes are additive (i.e., the volume of NaOH added after the second addition of the base is 0.50 + 1.00 = 1.50 mL, third time is 1.50 + 1.50 = 3.0 mL and so on ).



You are adding an acid to the buffer, calculate [NaOH] for each volume of acid added:

[NaOH]
=
0.100 M
x
V NaOH added

30.25 + VNaOH added


You will be subtracting this value from [HOAc] and adding it to [OAc-] in the ICE table:



Example: After addition of 0.50 mL of NaOH, [NaOH] = 0.00163

HOAc + H2O < = > H3O+ + OAc-]
Initial 0. 0462 - 0.00163 10-7 0.0513 + 0.00163
Change -x +x +x
Equilibrium 0.0446 - x x 0.0529 + x



Then into the equilibrium expression:

Ka = [H3O+][OAc-] = 1.76 x 10-5 = x (0. 0529 + x)
[HOAc] 0.0446 - x

5% of 0.0446 = 2.23 x 10-3

assume x is less than this value

1.76 x 10-5 = x (0. 0529)
0.0446

Solving for x = [H3O+]

[H3O+] = 1.48 x 10-5 < 2.23 x 10-3

pH = - log [H3O+] = 4.83 M

The sad part is that MANY years ago I could have helped you out. Now when I open my college books I think, WOW I used to know this stuff.

pH = pKa + log [A-]/[HA] :thinking:
this is correct

tep it sucks and i have to work with it every day