Acid Base Solutions - Answers
1. a) Buret is a long tube marked for volumes, usually in increments of 0.1 mL
b) Titrant is the solution in the buret.
c) Indicator is a weak acid or base that changes colour, depending on the pH of the solution.
d) End-point is the pH at which the indicator changes colour
pH = -logKa(indicator)
e) The stopcork is the valve at the bottom of a buret.
f) A standard solution is a solution of known concentration.
2. Given: 24.00 mL of 0.100 mol/L NaOH (aq) neutralizes 20.00 mL of HCl (aq)
Required: concentration of HCl (aq)
Solution: HCl (aq) + NaOH (aq) → H2O (l) + NaCl (aq)
Mole ratio 1 mol 1 mol
nNaOH = MNaOH X VNaOH = 0.100 mol/L X 0.0240 L = 0.002400 mol
nHCl = nNaOH = 0.002400 mol
Therefore: MHCl = nHCl = 0.002400 mol = 0.120 mol
VHCl 0.02000 L L
3. Solution: HC2H3O2 (aq) + NaOH (aq) → H2O (l) + NaC2H3O2 (aq)
a) mole ratio 1 mol 1 mol 1 mol 1 mol
nNaOH = 0.500 mol/L X 0.01340 L = 0.00670 mol
MHC2H3O2 = 0.00670 mol = 0.670 mol
0.0100 L L
b) 0.670 mol HC2H3O2 = 0.670 mol X 1 mL X 60.0 g
1 L 1000 mL 1.0 g 1 mol
= 40.2 g HC2H3O2 = 4.02 g HC2H3O2
1000 g vinegar 100 g vinegar
Therefore, vinegar is 4.02 % by weight acetic acid.
4. Solution: HC3H5O3 + NaOH → H2O + NaC3H5O3
mole ratio 1 mol 1 mol
nNaOH = 0.160 mol/L X 0.01835 L = 0.002936 mol
nHC3H5O2 = nNaOH = 2.94 X 10-3 mol
5. Solution: HC9H7O4 + KOH → KC9H7O4 + H2O
Mole ratio 1 mol 1 mol
nKOH = 0.0300 mol/L X 0.02940 L = 0.0008820 mol
nHC9H7O4 in 0.250 g of pain reliever = 8.820 X 10-4 mol
Therefore, the mass of pure HC9H7O4 in 0.250 g of pain reliever =
8.820 X 10-4 mol X 180.0 g = 1.588 X 10-1 g
1 mol
% of aspirin in pain reliever = 1.588 X 10-1 g X 100
0.250 g
= 63.5 %
Self-ionization of Water and pH – Answers
1. Kw = [H+][OH-] in neutral solution [H-]=[OH-]
2.4 X 10-14 = [H+]2
[H+] = 1.55 X 10-7 = [OH-]
pH = -log1.55 X 10-7 = 6.81
pOH = -log1.55 X 10-7 = 6.81
pH + pOH = 13.62 pKw = -logKw = -log2.4 X 10-14 = 13.62
Water at 37 C is neutral when pH = 6.81
2. [H3O+] = 0.01 mol/L Therefore, pH = -log0.010 = 2.0
3. [H3O+] = 0.0050 mol/L Therefore pH - -log0.0050 = 2.3
4. 6.0 g NaOH X 1 mol = 0.150 mol
1.00 L 40.0 g 1.00 L
[OH-] = 0.150 mol/L pOH = -log0.150 = 0.82
pH + pOH = 14.00 pH = 13.18 at 25 C
5. 0.837 g Ba(OH)2 = 8.37 g X 1 mol = 0.04886 mol
100 mL 1000 mL 171.3 g
Therefore 0.04886 mol Ba(OH)2 / L
But, Ba(OH)2 → Ba2+ + 2 OH-
Therefore 2 X 0.04886 mol OH/L [OH-] = 9.77 X 10-2 mol/L
pOH = -log9.77 X 10-2 = 1.01 pH = 12.99
6. [H3O+] = 1.9 X 10-5 mol/L pH = -log1.9 X 10-5 = 4.7
7. pH = -log1.4 X 10-5 = 4.9
Acid and Base Ionization Constants - Answers
1. Ka = [H3O+][C6H5CO2-]
[C6H5CO2H]
2. C6H5CO2- (aq) + H2O (l) ↔ C6H5CO2H (aq) + OH- (aq)
Kb = [C6H5CO2H][OH-]
[C6H5CO2-]
3. The lower the pKa, the stronger the acid. The stronger the acid, the weaker its conjugate base and vice versa. HF is a stronger acid than HCN, therefore the conjugate base of HF is weaker than the conjugate base of HCN. Therefore CN is the stronger base.
4. HF (aq) + H2O (l) ↔ H3O+ (aq) + F- (aq) Ka = 3.5 X 10-4
Initial 0.15 mol/L 0 0
Equilibrium 0.15-x x x
0.15_____ = 429 Ka = [H3O+][F-]
3.5 X 10-4 1 [HF]
3.5 X 10-4 = __x2__
0.15
x = 7.2 X 10-3
pH = -log7.2 X 10-3 = 2.1
5. HC2H3O2 (aq) + H2O (l) ↔ H3O+ (aq) + C2H3O2- (aq)
Initial 1.0 0 0
Equilibrium 1.0-x x x
Ka = 1.8 X 10-5
Ka = [H3O+][C2H3O2-]
[HC2H3O2]
1.8 X 10-5 = __x2__
1.0
x = 4.2 X 10-3
pH = -log4.2 X 10-3 = 2.4
Hydrolysis of Ions - Answers
1. Basic
2. Basic. H2C2O4 is a weak acid, therefore its conjugate base hydrolyzes producing a basic solution.
3. a) Acidic – (NH4)2, HC2H3O3
b) Basic – KF, KCN
c) Neutral – NaI, CsNO3, KBr
CN- (aq) + H2O (l) ↔ HCN (aq) + OH-
Initial (mol/L) 0.20 0 0
Equilibrium 0.20-x x x
Kb = Kw = 1.0 X 10-14 = 2.0 X 10-5
Ka 4.9 X 10-10
Kb = [HCN-][OH-]
[CN-]
2.0 X 10-5 = __x2_
0.20
x = 2.0 X 10-3 = [OH-]
pOH = 2.7 pH = 11.3
NO2- (aq) + H2O (l) ↔ HNO2 (aq) + OH- (aq)
Initial (mol/L) 0.10 0 0
Equilibrium 0.10-x x x
Kb = [HNO2][OH-]
[NO2-]
2.2 X 10-11 = __x2__
0.10
x = 1.5 X 10-6 = [OH-]
pOH = 5.8
but pH + pOH = 14
pH = 8.2
4. CO32- (aq) + H2O (l) ↔ HCO3- (aq) + OH- (aq)
Initial (mol/L) x 0 0
Equilibrium x-5 X 10-3 5 X 10-3 5 X 10-3
Kb = Kw = 1.0 X 10-14 = 2.1 X 10-4
Ka 4.7 X 10-11
pH = 11.70 pOH = 2.30 [OH-] = 5.0 X 10-3
Kb = [HCO3-][OH-]
[CO32-]
2.1 X 10-4 = (5.0 X 10-3)2
x
x = 0.12 mol/L
but 1 mol of CO32- per 1 mol Na2CO3∙10H2O. Therefore mass of Na2CO3∙10H2O
= 0.12 mol X 286.0 g = 34 g
1 L 1 mol 1 L
6. pH = 10.00 pOH = 4.00 [OH-] = 10-4.00 = 1.00 X 10-4 mol/L
SO3 2- (aq) + H2O (l) ↔ HSO3- (aq) + OH- (aq)
Initial x 0 0
Equilibrium x-1 X 10-4 1 X 10-4 1 X 10-4
Kb = Kw = 1.00 X 10-14 = 1.5 X 10-7
Ka 6.6 X 10-8
Kb = [HSO3-][OH-]
[SO32-]
1.5 X 10-7 = (1.00 X 10-4)2
x
x = 6.7 X 10-2
6.7 X 10-2 mol/L of SO32-
3.33 X 10-2 mol X 252.1 g = 8.4 g/0.5 L
0.5 L 1 mol
7. a) HSO4- (aq) + H2O (l) ↔ H3O+ (aq) + SO42- (aq)
Ka = [H3O+][SO42-] = 1.0 X 10-2
[HSO4-]
b) HSO4- (aq) + H2O (l) ↔ H3O+ (aq) + SO42- (aq)
Initial 0.010 0 0
Eqm. 0.010-x x x
1.0 X 10-2 = ___x2__
0.010-x
1.0 X 10-4 – 1.0 X 10-2 x = x2
x2 + 1.0 X 10-2 x – 1.0 X 10-4 = 0
x = 1.0 X 10-2 +/- √1.0 X 10-4 + 1.0 X 10-4
2
= -1.0 X 10-2 +/- 1.4 X 10-2
2
= -1.2 X 10-2 or 0.2 X 10-2
x = 0.2 X 10-2
[H3O+] = 2.0 X 10-3 mol/L
c) Using simplification (0.010 – x) = 0.010
1.0 X 10-2 = __x2__
0.010
x = 1.0 X 10-2
[H3O+] = 1.0 X 10-2
d) % ionization = 2.0 X 10-3 X 100 = 20 %
0.010
e) % ionization = 1.0 X 10-2 X 100 = 100%
0.010