1. pH = -log[H+] pH = -log[H+] pOH = -log[OH-] pH = -log[H+]
Acidic/Basic
5.55
9.17
1.12 x10-5 M
6.28 x 10-1 M
acidic
2.82 x 10-6
3.55 x 10-9
8.45
acidic
1.48 x 10-5
6.76 x 10-10
4.83
8.93 x 10-10
basic
9.05
4.95
1.59 x 10-14
0.20
13.8
very acidic
pH = -log[H+]
pH = -log[H+]
pOH = -log[OH-]
pH = -log[H+]
pH = -log[8.93 x 10-10]
pH = -log[6.28 x 10-1]
9.17 = -log[OH-]
5.55 = -log[H+]
[H+][OH-] = 1 x 10-14
[H+][OH-] = 1 x 10-14
[H+][OH-] = 1 x 10-14
[H+][OH-] = 1 x 10-14
pOH = -log[OH-]
pOH = -log[OH-]
[OH-] =
[H+] =
[H+][1.12 x 10-5] = 1 x 10-14
[6.28 x 10-1][OH-] = 1 x 10-14
[2.82 x 10-6][OH-] = 1 x 10-14
[H+][6.76 x 10-10] = 1 x 10-14
pH = -log[1.59 x 10-15]
pH = -log[1.12 x 10-5]
pH + pOH = 14
pH + pOH = 14
[H+] = 8.93 x 10-10
[H+] = 3.55 x 10-9
[H+] = 1.48 x 10-5
[OH-] = 1.59 x 10-14

2. acid = HC2H3O2 , conjugate base = C2H3O2- ions
2. a) HC2H3O2 + H2O → H3O+ + C2H3O2-
acid = HC2H3O2 , conjugate base = C2H3O2- ions
base = H2O , conjugate acid = H3O+ ions
b) OH- + HCO3- → CO32- + H2O
acid = HCO3- ions, conjugate base = CO32- ions
base = OH- ions, conjugate acid = H2O
c) SO HF → HSO3- + F-
acid = HF, conjugate base = F- ions
base = SO32- ions, conjugate acid = HSO3- ions

3. a) @ equivalence point there is 27.4 mL of 0.135 M NaOH . . . .
(0.135)(0.0247) = 3.7 x 10-3 mols NaOH since HA is monoprotic,
stoichometric equivalent amounts means there will be
3.7 x 10-3 mols of acid
HA (aq) + NaOH (aq) → NaA (aq) + H2O (l)
b) Molecular weight is old people speak for molar mass
molar mass means grams per mol
184.4 g/mol

4. when pH =
10.6 milliliters of M NaOH has been added
(0.0106L)(0.135M) = mole OH- has been added
thus mol acid has been removed
3.7 x = mol acid remaining
pH = 5.65
pH = - log[H+]
log-1(-5.65)
2.23 x 10-6 M

5. HC2H3O2 (aq) + K+ (aq) + OH- (aq) → K+ (aq) + C2H3O2- (aq) + H2O (l)
4. a) HC2H3O2 (aq) + KOH (aq) → KC2H3O2 (aq) + H2O (l)
HC2H3O2 (aq) + K+ (aq) + OH- (aq) → K+ (aq) + C2H3O2- (aq) + H2O (l)
HC2H3O2 (aq) + OH- (aq) → C2H3O2- (aq) + H2O (l)
b) NH3 (aq) + HNO3 (aq) → NH4NO3 (aq)
NH3 (aq) + H+ (aq) + NO3- (aq) → NH4+ (aq) + NO3- (aq)
b) NH3 (aq) + H+ (aq) → NH4+ (aq)
c) 2HClO3 (aq) + Sr(OH)2 (aq) → Sr(ClO3)2 (aq) + 2H2O (l)
2H+(aq) + 2ClO3-(aq) + Sr2+(aq) + 2OH-(aq) → Sr2+(aq) + 2ClO3-(aq) + 2H2O(l)
H+(aq) + OH-(aq) → H2O(l)
d) 2HIO3 (aq) + Cu(OH)2 (aq) → Cu(IO3)2 (aq) + 2H2O (l)
2HIO3 (aq) + Cu(OH)2 (aq) → Cu2+ (aq) + 2IO3- (aq) + 2H2O (l)
2HIO3 (aq) + Cu(OH)2 (aq) → Cu2+ (aq) + 2IO3- (aq) + 2H2O (l)

6. a) 1.15 g x 0.30 = 0.375 g HCl in 1 mL of solution
therefore g HCl in 1000 mL of solution
therefore mol HCl in 1000 mL of solution
9.46 M
106 mL
b) MaVa = MbVb (9.46)Va = (0.2)(5) c)
mol OH-
therefore mol H+ in mL HCl solution M

7. Lewis acid = BF3 (electron pair acceptor)
Lewis base = NH3 (electron pair donor)
a) HNO3 (aq) → H+ (aq) + NO3- (aq) b) HF (aq) ⇄ H+ (aq) + F- (aq)
c) HBr (aq) → H+ (aq) + Br- (aq) d) HC2H3O2 (aq) ⇄ H+ (aq) + C2H3O2- (aq)
e) H2SO4 (aq) → H+ (aq) + HSO4- (aq) f) H2CO3 (aq) ⇆ H+ (aq) + HCO3- (aq)
HSO4- (aq) ⇆ H+ (aq) + SO42- (aq) HCO3- (aq) ⇆ H+ (aq) + CO32- (aq)
g) H3PO4 (aq) ⇆ H+ (aq) + H2PO4- (aq)
H2PO4- (aq) ⇆ H+ (aq) + HPO42- (aq)
HPO42- (aq) ⇆ H+ (aq) + PO43- (aq)
HSO4- (aq), HCO3- (aq), H2PO4- (aq) and HPO42- (aq)

8. (3.3 × 10-5)( ) = 4.39 × 10-7
= [H+] pH = 6.36
(1.5 × 10-3)( ) = 6.75 × 10-6
= [OH-] pOH =
thus pH = 8.83
c) pH = [H+] = log-1(-4.5) =
3.162 × 10-5 is 1.33 % of what?
3.162 × 10-5 = (0.0133)(?) M
10) a) b) c) d) 10.88

9. 11. a is neutral KOH and HCl are both strong
b is acidic both are weak but the Ka for oxalic acid is larger than the Kb for ammonia
c is acidic Cu(OH)2 is weak and HCl is strong
d is basic CsOH is strong and H2Cr2O7 is weak
e is acidic both are weak but the Ka for acetic acid is larger than the Kb for copper(II) hydroxide
HCl (aq) + NaOH (aq) → NaCl (aq) + H2O (l)
25 mL of 0.10 M HCl mol HCl
therefore mol NaOH in 50 mL;
0.05 M NaOH

10. 13. same equivalence point; less NaOH needed to reach equivalence point
14. higher equivalence point; less NaOH needed to reach equivalence point
15. lower equivalence point; more NH3 needed to reach equivalence point