A B C D E F 1 Solution pH Volume (mL) [H + ] (M) [OH ] (M) Volume for pH=7 (L) 2
ID: 1032349 • Letter: A
Question
A
B
C
D
E
F
1
Solution
pH
Volume (mL)
[H+] (M)
[OH] (M)
Volume for pH=7 (L)
2
0.10 M Acetic Acid
3.2
20
3
Baking Soda
8
50
4
0.10 M HCl
1.75
20
5
0.10 M NaOH
12.27
20
6
Ammonium Chloride
6.1
20
7
Bleach
12.7
20
8
Sodium Acetate
7.56
50
9
Vinegar
3.05
20
10
Apple Juice
4.01
20
11
Coke
2.53
20
12
Sodium Chloride
7.75
20
Perform the following calculations in Columns C and D by entering expressions into the appropriate cells in Excel.
Column D - take the measured pH values from Column B and calculate the concentration of H+ in molarity.
Column E - take the calculated concentration of H+ from Column D and calculate the concentration of OH in molarity
Column F - calculate the volume you would have to dilute your solution to for it to have a pH of 7 using the formula M1V1 = M2V2. (Hint: for acidic solutions you will need to use [H+] and for basic solutions you will need to use [OH])
A
B
C
D
E
F
1
Solution
pH
Volume (mL)
[H+] (M)
[OH] (M)
Volume for pH=7 (L)
2
0.10 M Acetic Acid
3.2
20
3
Baking Soda
8
50
4
0.10 M HCl
1.75
20
5
0.10 M NaOH
12.27
20
6
Ammonium Chloride
6.1
20
7
Bleach
12.7
20
8
Sodium Acetate
7.56
50
9
Vinegar
3.05
20
10
Apple Juice
4.01
20
11
Coke
2.53
20
12
Sodium Chloride
7.75
20
Explanation / Answer
2. 0.1 M acetic acid
pH = -log[H+] = 3.2
[H+] = 6.31 x 10^-4 M
[OH-] = 1 x 10^-14/6.31 x 10^-4 = 1.58 x 10^-11 M
initial [H+] = 6.31 x 10^-4 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/6.31 x 10^-4 M = 1.6 x 10^-4 L of 0.1 M acetic acid
--
3. 0.1 M baking soda
pH = -log[H+] = 8.0
[H+] = 1 x 10^-8 M
[OH-] = 1 x 10^-14/1 x 10^-8 = 1 x 10^-6 M
initial [OH-] = 1 x 10^-6 M
final [OH-] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/1 x 10^-6 M = 0.1 L
--
4. 0.1 M HCl
pH = -log[H+] = 1.75
[H+] = 0.018 M
[OH-] = 1 x 10^-14/0.018 = 5.62 x 10^-13 M
initial [H+] = 0.018 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/0.018 M = 5.55 x 10^-6 L
--
5. 0.1 M NaOH
pH = -log[H+] = 12.27
[H+] = 5.37 x 10^-13 M
[OH-] = 1 x 10^-14/5.37 x 10^-13 = 0.0186 M
initial [OH-] = 0.0186 M
final [OH-] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/0.0186 M = 5.38 x 10^-6 L
--
6. 0.1 M NH4Cl
pH = -log[H+] = 6.1
[H+] = 7.94 x 10^-7 M
[OH-] = 1 x 10^-14/7.94 x 10^-7 = 1.26 x 10^-8 M
initial [H+] = 7.94 x 10^-7 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/7.94 x 10^-7 M = 0.126 L
--
7. 0.1 M bleach
pH = -log[H+] = 12.7
[H+] = 2 x 10^-13 M
[OH-] = 1 x 10^-14/2 x 10^-13 = 0.050 M
initial [OH-] = 0.050 M
final [OH-] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/0.050 M = 2 x 10^-6 L
--
8. 0.1 M sodium acetate
pH = -log[H+] = 7.56
[H+] = 2.75 x 10^-8 M
[OH-] = 1 x 10^-14/2.75 x 10^-8 = 3.63 x 10^-7 M
initial [OH-] = 3.63 x 10^-7 M
final [OH-] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/3.63 x 10^-7 M = 0.275 L
--
9. 0.1 M vinegar
pH = -log[H+] = 3.05
[H+] = 8.91 x 10^-4 M
[OH-] = 1 x 10^-14/8.91 x 10^-4 = 1.12 x 10^-11 M
initial [H+] = 8.91 x 10^-4 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/8.91 x 10^-4 M = 1.12 x 10^-4 L
--
10. 0.1 M apple juice
pH = -log[H+] = 4.01
[H+] = 9.8 x 10^-5 M
[OH-] = 1 x 10^-14/9.8 x 10^-5 = 1.02 x 10^-10 M
initial [H+] = 9.8 x 10^-5 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/9.8 x 10^-10 M = 1.02 x 10^-3 L
--
11. 0.1 M coke
pH = -log[H+] = 2.53
[H+] = 2.95 x 10^-3 M
[OH-] = 1 x 10^-14/2.95 x 10^-3 = 3.39 x 10^-12 M
initial [H+] = 2.95 x 10^-3 M
final [H+] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/2.95 x 10^-3 M = 3.4 x 10^-5 L
--
12. 0.1 M NaCl
pH = -log[H+] = 7.75
[H+] = 1.8 x 10^-8 M
[OH-] = 1 x 10^-14/1.8 x 10^-8 = 5.62 x 10^-7 M
initial [OH-] = 5.62 x 10^-7 M
final [OH-] needed = 1 x 10^-7 M
let 1.0 L be the final volume of solution
then,
volume for pH 7 = 1 x 10^-7 M x 1 L/5.62 x 10^-7 M = 0.180 L
--
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