Calculate pH using 40.0 mL 0.10M HAc and:
(a) 0 mL NaOH
(b) 20.00 mL NaOH
(c) 30.00 mL NaOH
(d) 40.00 mL NaOH
(e) 50.00 mL NaOH
Calculate pH using 40.0 mL 0.10M HAc and:
(a) 0 mL NaOH
(b) 20.00 mL NaOH
(c) 30.00 mL NaOH
(d) 40.00 mL NaOH
(e) 50.00 mL NaOH
(a) 0.100 M HAc
HAc 
H+
+ Ac-
.1-x x
x Ka = x2/(.1) = 1.8x10-5
x = [H+] = 1.34x10-3 pH = 2.87
(b) HAc:
mol = n = MxL = (0.100)(0.0400) = 0.00400 mol
NaOH:
n = MxL = (0.100)(0.0200) = 0.00200 mol
HAc
in excess by 0.00200 mol, 0.00200 mol Ac-1 produced!
total
volume = 60.00 mL = 0.06000 L
We
now have buffer with HAc, H+1, and Ac-1
[HAc] = 0.00200 mol/0.0600L = 0.333M
[H+1] = ?
[Ac-1] = 0.00200 mol/0.0600L = 0.333M
To attain equilibrium, shift takes place to to the right.
HAc
H+ + Ac-
.333-x
x .333+x
After
approximation, Ka = [H+1] = 1.8x10-5
pH = 4.74
(c) HAc:
n = MxL = (0.100)(0.0400) = 0.00400 mol
NaOH:
n = MxL = (0.100)(0.0300) = 0.00300 mol
HAc
in excess by 0.00100 mol, 0.00300 mol Ac-1 produced!
total
volume = 70.00 mL = 0.07000 L
We
now have buffer with HAc, H+1, and Ac-1
[HAc] = 0.00100 mol/0.0700L = 0.0143M
[H+1] = ?
[Ac-1] = 0.00300 mol/0.0700L = 0.0429M
To attain equilibrium, shift takes place to to the right.
HAc H+
+ Ac-
.0143-x
x .0429+x
After
approximation, Ka = x(.0429)/(0.0143) = 1.8x10-5
x = [H+1] = 6.0x10-6
pH = 5.22
(d) HAc:
n = MxL = (0.100)(0.0400) = 0.00400 mol
NaOH:
n = MxL = (0.100)(0.0400) = 0.00400 mol
[NaAc]
= 0.00400mol/0.0800L = 0.0500M
Solution
not neutral because Ac-1 undergoes
hydrolysis
Ac-1
HAc + OH-1
0.050 x
x
Kb
= [HAc][OH-1]/[Ac-1]
= x2/0.050 = 5.6x10-10
x =
[OH-1] = 5.3x10-6
pOH = 5.28 pH =8.72
(e) HAc:
n = MxL = (0.100)(0.0400) = 0.00400 mol
NaOH:
n = MxL = (0.100)(0.0500) = 0.00500 mol
NaOH
in excess by 0.00100 mol, volume = 90.00 mL = 0.09000 L
[OH-1]
= 0.00100mol/0.09000L = 0.0111 M pOH = 1.95 pH= 12.05
