Help these problems from Acids, Bases, Buffers and Salts lab. a) Strong Acids an
ID: 508763 • Letter: H
Question
Help these problems from Acids, Bases, Buffers and Salts lab.
a) Strong Acids and Bases
Molarity (HCl)
0.1M
0.01M
0.001M
pH
1.30
2.14
3.03
pH
Molarity (NaOH)
0.1M
0.01M
0.001M
pH
12.91
11.95
10.90
pH
b) Weak Acids and Bases
Molarity (HC2H3O2)
1.0M
0.1M
0.01M
pH
2.43
2.90
3.03
pH
[H3O+]
[CH3O2-]
[HC2H3O2]
Ka
Average
pKa
Molarity (NH3)
1.0M
0.1M
0.01M
pH
11.79
11.12
10.65
pH
[OH-]
[NH4+]
[NH3]
Kb
Average
pKb
1. How do the pH measurements of 0.1M NaOH and 0.1M NH3 show that NaOH is a stronger base than NH3 ?
2. Use the pH measurement you made for each HC2H3O2 solution to calculate a value of Ka and pKa for HC2H3O2. Are all 3 of the pKa values you calculated similar?
3. Use the pH measurement you made for each NH3 solution to calculate a value of Kb and pKb for NH3. Are all 3 of the pKb values you calculated similar?
c) Buffer Solutions
Molarity (Buffer)
1.0M
0.1M
0.01M
pH before acid
4.68
4.63
4.62
pH after base
4.62
4.17
2.27
pH
1. How did diluting the buffer affect the pH of the buffer solution itself?
2. Calculate the change in pH produced by the addition of the HCl to each buffer solution. Why are the results for each of the three solutions different? (They should be)
d) Salt solutions (using pH paper)
Sodium Carbonate = pH 10
Aluminum Chloride = pH 4
Ammonium Carbonate = pH 8
Ammonium Chloride = pH 6
1. Examine the pH of solutions of Na2CO3, NH4Cl, (NH4)2CO3 and AlCl3.
2. Based on the pH you measured for the solution of (NH4)2CO3 is NH4+ a stronger acid or is CO32- a stronger base? Explain.
Molarity (HCl)
0.1M
0.01M
0.001M
pH
1.30
2.14
3.03
pH
Explanation / Answer
1). The pH scale measures how acidic or basic a substance is. The pH scale ranges from 0 to 14. A pH of 7 is neutral. A pH less than 7 is acidic. A pH greater than 7 is basic. NaOH is a stronger base than NH3 because pH of 0.1M NaOH is 12.91 which is higher than the pH of 0.1M NH3 which is 11.79.
2) The dissociation equation for the acetic acid is
CH3COOH H+ + CH3COO¯
Ka = ( [H+] [CH3COO ¯] ) / [CH3COOH]
pH = -log [H+], therefore [H+] = 10¯Ph
pH = 2.43
[CH3COOH] = 1M
[H+] = 10¯2.43 = 3.72 x 10¯3 M
From the dissociation equation, we know there is a 1:1 molar ratio between [H+] and [CH3COO¯]. Therefore:
[CH3COO¯] = 3.72 x 10¯3 M
Ka = 3.72 x 10¯3 M x 3.72 x 10¯3 M ¸ 1M = 1.38 x 10-5
PKa = -log Ka = 4.86
b) pH = 2.90
[CH3COOH] = 0.1M
[H+] = 10¯2.90 = 1.26 x 10¯3 M
[CH3COO¯] = 1.26 x 10¯3 M
Ka = 1.26 x 10¯3 M x 1.26 x 10¯3 M ¸ 0. 1M = 1.59 x 10-5
PKa = -log Ka = 4.80
c) pH = 3.03
[CH3COOH] = 0.01M
[H+] = 10¯3.03 = 9.33 x 10¯4 M
[CH3COO¯] = 9.33 x 10¯4 M
Ka = 9.33 x 10¯4 M x 9.33 x 10¯4 M ¸ 0. 01M = 8.7 x 10-5
PKa = -log Ka = 4.06
3). the ionizaton equation for NH3.
NH3 + H2O NH3+ + OH¯
Kb = ( [NH3+] [OH¯] ) / [NH3]
We will use the pH to calculate the [OH¯]. We know pH = -log [H+], therefore [H+] = 10¯pH
pH = 11.79
[NH3] = 1M
[H+] = 10¯11.79 = 1.62 x 10¯12 M
Kw = [H+] [OH¯] = 1 x 10-14
[OH¯] = 1 x 10-14 ¸ 1.62 x 10¯12 M = 6.17 x 10-3 M
Kb = ( [NH3+] [OH¯] ) / [NH3] = 6.17 x 10-3 x 6.17 x 10-3 ¸ 1 = 38.07 x 10-6
pKb = -log 38.07 x 10-6 = 4.42
b) pH = 11.12
[NH3] = 0.1M
[H+] = 10¯11.12 = 7.59 x 10¯12 M
Kw = [H+] [OH¯] = 1 x 10-14
[OH¯] = 1 x 10-14 ¸ 7.59 x 10¯12 M = 1.31 x 10-3 M
Kb = ( [NH3+] [OH¯] ) / [NH3] = 1.31 x 10-3 x 1.31 x 10-3 ¸ 0.1 = 17.16 x 10-6
pKb = -log 17.16 x 10-6 = 4.76
c) pH = 10.65
[NH3] = 0.01M
[H+] = 10¯10.65 = 2.24 x 10¯11 M
Kw = [H+] [OH¯] = 1 x 10-14
[OH¯] = 1 x 10-14 ¸ 2.24 x 10¯11 M = 0.45 x 10-3 M
Kb = ( [NH3+] [OH¯] ) / [NH3] = 0.45 x 10-3 x 0.45 x 10-3 ¸ 0.01 = 19.93 x 10-6
pKb = -log 19.93 x 10-6 = 4.70
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