Answer:
a) <u>2.79 </u>
b) <u>4.75</u>
c) 6.59
d) 8.81
e) 11.03
f) 12.42
Explanation:
Step 1: Data given
Volume of acetic acid = 35.0 mL
Molarity of acetic acid = 0.150 M
Molarity of NaOH = 0.150 M
Step 2: pH after adding 0 mL of NaOH
HA + H2O ⇆ A- + H3O+
Initial concentration
[HA] = 0.150 M
[A-] = 0M
[H3O+] = 0M
Concentration at the equilibrium
[HA] = 0.150 - x M
[A-] = xM
[H3O+] = xM
Ka = [H3O+][A-] / [HA] = x²/(0.150 -x) = 1.76*10^-5
0.150 >>> x
We can write it like this
x²/(0.150) = 1.76*10^-5
x² =0.00000264
x = 0.00162
[H3O+] = 0.00162 M
pH = -log [H3O+] = -log (0.00162) = <u>2.79 </u>
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Step 3: pH after adding 17.5 mL of NaOH
This is the point of half-neutralization.
pH = pKa = -log (1.76*10^-5- 4.75
Step 4: pH after adding 34.5 mL of NaOH
HA + OH- ⇆ A- + H2O
Moles acetic acid = molarity * volume
Moles acetic acid = 0.150 M * 0.035 L
Moles acetic acid = 0.00525 moles
Moles NaOH = 0.150 M *0.0345 L
Moles NaOH = 0.005175 moles
Initial moles
[HA] = 0.00525 moles
[OH-] = 0.005175 moles
[A-] = 0 moles
[H2O] = 0 moles
Moles at the equilibrium
[HA] = 0.00525 - 0.005175 = 0.000075 moles
[OH-] = 0.005175 - 0.005175 = 0 moles
[A-] = 0.005175 moles
pH = pKa + log([A-]/[HA]
pH = 4.75 +log(0.005175/0.000075)
pH = 4.75 + 1.84 = <u>6.59</u>
Step 4: pH after adding 35.0 mL of NaOH
HA + OH- ⇆ A- + H2O
Moles acetic acid = molarity * volume
Moles acetic acid = 0.150 M * 0.035 L
Moles acetic acid = 0.00525 moles
Moles NaOH = 0.150 M *0.035 L
Moles NaOH = 0.00525 moles
Initial moles
[HA] = 0.00525 moles
[OH-] = 0.00525 moles
[A-] = 0 moles
[H2O] = 0 moles
Moles at the equilibrium
[HA] = 0.00525 - 0.00525 = 0 moles
[OH-] = 0.00525 - 0.00525 = 0 moles
[A-] = 0.00525 moles
We have a solution of 0.00525 mol of A
- in 70.0 mL
[A-] = 0.00525 moles / 0.070 L = 0.075 M
A- + H20 ⇆ HA + OH-
Initial moles A- = 0.0750 moles
initial moles HA = 0 moles
Initial moles OH- = 0 moles
Moles A- at the equilibrium = 0.0750 - x
Moles HA at the equilibrium = x
Moles OH- at the equilibrium = x
Kb = Kw/Ka = 10^-14 / 1.76*10^-5
Kb = 5.68 *10^-10
Kb = [HA][OH-] / [A-] = x² / 0.0750 - x = 5.68*10^-10
Since 0.0750 >>> x we can write
5.68*10^-10 = x²/0.0750
x = 6.53 *10^-6
[OH-] = x = 6.53 * 10^-6 M
pOH = - log (6.53*10^-6) = 5.19
pH = 14 - pOH = 14 - 5.19 =<u> 8.81</u>
Step 5: pH after adding 35.5 mL of NaOH
We add an excess moles of NaOH
We have an excess of 0.5 mL NaOH
Moles excess = 0.0005 L *0.150 M = 0.000075 moles
[OH-] = 0.000075 moles / (0.035+0.0355L)
[OH-] = 1.06 * 10^-3 M
pOH = -log( 1.06 *10^-3)
pOH = 2.97
pH = 14 -2.97 = <u>11.03</u>
Step 6: pH after adding 50.0 mL of NaOH
We have an excess of 15.0 mL NaOH
Moles excess = 0.015 L *0.150 M = 0.00225 moles
[OH-] = 0.00225 moles / (0.035+0.050L)
[OH-] = 0.0265 M
pOH = -log( 0.0265M)
pOH = 1.58
pH = 14 -1.58 = <u>12.42</u>
