Name: _________________________________________
Date:____________________________
Optional!!! Homework 11 – do not turn in!!!
1.) Given the following [H+1] concentrations, calculate pH, pOH, and [OH-1]. Are the solutions acidic or
basic?
a. [H+1] = 2.00 x 10-3 M
pH = -log(H+1)
pH = 2.70 ACIDIC
pH = -log(2.00 x 10-3) = 2.70
pOH = 11.30
14.00 = pH + pOH
14.00 = 2.70 + pOH
[OH-1] = 5.01 x 10-12 M
14.00 – 2.70 = pOH = 11.30
OR!!!
pOH = -log[OH-1]
11.30 = -log[OH-1]
1.00 x 10-14 = [H+1][OH-1]
[OH-1] = 1.00 x 10-14/2.00 x 10-3 = 5.00 x 10-12 M OH-1
10-11.30 = 5.01 x 10-12 M OH-1
pOH = -log[OH-1] = -log(5.00 x 10-12) = 11.30
b. [H+1] = 7.67 x 10-3 M
pH = -log (7.67 x 10-3) = 2.11
pH = 2.11 ACIDIC
14.00 = pH + pOH
14.00 = 2.11 + pOH
pOH = 11.88
pOH = 11.88
pOH = -log[OH-1]
11.88 = -log [OH-1]
OH-1 = 10-11.88 = 1.32 x 10-12 M
[OH-1] = 1.32 x 10-12 M
2.) Given the following pOH values, calculate your concentration of [H+1], [OH-1], and pH (don’t forget
units!!)
pH = 12.45
a. pOH = 1.55
[H+1] = 3.55 x 10-13 M
14.00 = pH + pOH
14.00 = pH + 1.55
[OH-1] = 2.82 x 10-2 M
pH = 12.45
pH = -log[H+1]
pOH= -log[OH-1]
10-12.45 = 3.55 x 10-13 M H+1
10-1.55 = 2.82 x 10-2 M OH-1
_______________________________________________________________________________________
b. pOH = 4.25
pH = 9.75
14.00 = pH + pOH
14.00 = pH + 4.25
[H+1] = 1.78 x 10-10 M
[OH-1] = 5.62 x 10-5 M
pH = 9.75
pH = -log[H+1]
pOH= -log[OH-1]
10-9.75 = 1.78 x 10-10 M H+1
10-4.25 = 5.62 x 10-5 M OH-1
Another way to check on some of the answers is to take the calculated [H+1] x [OH-1] and make
sure that you get something close to 1.00 x 10-14!!
For example, 2a has an [OH-1] = 2.82 x 10-2 M and an [H+1] = 3.55 x 10-13
2.82 x 10-2 * 3.55 x 10-13 = 1.00 x 10-14 !
3.) A 0.0500 M acetic acid solution is tested with a pH probe and found to have a pH of 3.03 at 25oC.
Use this information to calculate Ka for acetic acid if acetic acid dissociates into H3O+1 and the
acetate ion in water.
HAc
0.0500
-x
0.0500 - x
+
H2 O
X
X
X
↔
pH = -log(H+1)
3.03 = -log(H+1)
10-3.03 = [H+1] = 9.33 x 10-4 M
[H3O+1] = 9.33 x 10-4 M = x
[H3O+1]eq = 9.33 x 10-4 M
[Ac-1]eq = 9.33 x 10-4 M
[HAc] = 0.0500 – 0.000933 = 0.0491 M
[H 3 O 1 ][Ac 1 ] [9.33 x 10 -4 ][9.33 x 10 -4 ]
Ka =
=
=
[HAc]
[0.0491]
1.78 x 10-5
Ac-1
0
+x
x
+
H3O+1
0
+x
x
4.) Ascorbic acid (formula does not matter at this time!), better known as vitamin C, is found in citrus
fruits. The acid has a Ka value of 1.0 x 10-5. Calculate the concentration of ascorbic acid, the
ascorbate ion, and the H+1 (H3O+1) at equilibrium if you have a solution of 0.050 M ascorbic acid.
Vit C
0.050
-x
0.050 - x
Ka =
+
H2 O
X
X
X
↔
[H 3 O 1 ][C 1 ]
[x][x]
=
= 1.0 x 10-5
[0.050 - x]
[VitC]
Assume x is small!! Such that 0.050 –x ≈ 0.050
1.0 x 10-5 =
[x][x]
[0.050]
5.0 x 10-7 = x2
x = 7.07 x 10-4 M
assumption check!!
7.07 x 10 4
x 100 = 1.41 % which is less than 5% 
0.050
Vitamin C = 0.050 – 0.00071 ≈ 0.050 M
C-1 = 7.1 x 10-4 M
H3O+1 = 7.1 x 10-4 M
***Extra*** pH = -log(H+1) = -log(7.1 x 10-4) = 3.1
C-1
0
+x
x
+
H3O+1
0
+x
x
5.) Calculate the concentrations of H+1, HC2O4-1, and C2O4-2 in a 0.10 M H2C2O4 solution. Ka1 = 5.90 x
10-2 and Ka2 = 6.40 x 10-5. (HINT: you will need two separate ICE tables and TWO separate
calculations. The first will use Ka1 and the second will use Ka2)
H2C2O4
0.10
-x
0.10 - x
+
↔
H2 O
X
X
X
H1C2O4-1
0
+x
x
-1
Ka1 =
[H 3 O 1 ][HC2 O 4 ]
[x][x]
=
= 5.90 x 10-2
[H 2 C 2 O 4 ]
[0.10 - x]
Assume x is small!! Such that 0.10 –x ≈ 0.10
[x][x]
= 5.90 x 10-2
[0.10]
x2 = 5.9 x 10-3
x = 0.077 M
assumption check!!
0.077
x 100 = 76.8 % which NOT less than 5% 
0.10
Must use quadratic!!
[x][x]
= 5.90 x 10-2
[0.10 - x]
5.9 x 10-3 – 5.9 x 10-2x = x2
0 = x2 + 5.9 x 10-2 x – 5.9 x 10-3
x=
x=
x=
 5.9x102  (5.9x102 ) 2  (4(1)(-5.9x10-3 ))
 b  b 2  4ac
=
2a
2(1)
 5.9x102  (0.00348) (0.0236)
2
=
 5.9x102  (0.0271)
2
 5.9x102  0.1646
 5.9x102  0.1646
or x =
2
2
x =0.053
or x = - 0.11
Your concentration (which is x!) cannot be negative!
+
H3O+1
0
+x
x
So x = 0.053 M
[H2C2O4] = 0.10 – 0.053 = 0.05 M
[HC2O4-1] = 0.053 M
[H3O+1] = 0.053 M
Because we have starting concentrations of HC2O4-1 AND of H3O+1 already present from the parent acid
dissociation, we MUST put those starting amounts in our ICE table
HC2O4-1
0.053
-x
0.053 - x
+
H2 O
X
X
X
-2
Ka2 =
[H 3 O 1 ][C 2 O 4 ]
-1
[H 1 C 2 O 4 ]
=
↔
C2O4-2
0
+x
x
+
H3O+1
0.053
+x
0.053 + x
[x  0.053][x]
= 6.40 x 10-5
[0.053 - x]
Let’s assume that x is small – such that 0.053 –x ≈ 0.053 AND 0.053 + x ≈ 0.053
[ 0.053][x]
= 6.40 x 10-5
[0.053]
Look at how that math really simplifies things!!
x = 6.40 x 10-5 M
Check that assumption!!
[ 6.40 x10-5 ]
x100 = 0.12% which is less than 5% so  assumption!!
[0.053]
[C2O4-2] = 6.40 x 10-5 M
[H3O+1] = 0.053 + 6.40 x 10-5 = 0.053 M
[HC2O4-1] 0.053 – 6.40 x 10-5 = 0.053 M
And from our initial calculation H2C2O4 = 0.05 M
6.) List phosphoric acid and all of its conjugate base ions in order of increasing acidity and increasing
basicity (HINT: think about what it means for a species to be an acid!! Think about what it means
for a species to be a base!)
H3PO4  H2PO4-1 + H+1
H2PO4-1  HPO4-2 + H+1
HPO4-2  PO4-3 + H+1
H3PO4 > H2PO4-1 > HPO4-2 for acid strength (PO4-3 is NOT an acid as it doesn’t have any H+1 to
lose!)
PO4-3 > HPO4-2 > H2PO4-1 > H3PO4 for base strength (H3PO4 could be a base and accept
another H+1)