Chem 102 Recitation
pH, Weak Acid & Base
---Dr. Q. Wang
Mar 26, 2015
[H+], [OH-], and pH
At 25 C, for any aqueous solutions,
[H3O+] and [H+] are the same thing.
[H+] × [OH-] = 1.0 × 10-14
-log[H+] = pH, [H+] = 10-pH
-log[OH-] = pOH, [OH-] = 10-pOH
pH + pOH =14.00
pH < 7.0, acidic; pH > 7.0, basic.
[H+], [OH-], and pH
[H+] = 10-pH = 10-3.15 = 7.1×10-4 M
[H+] × [OH-] = 1.0 × 10-14
[OH-] = (1.0 × 10-14)/ [H+]
= (1.0 × 10-14)/ 7.1×10-4
= 1.4 × 10-11 M
pH < 7.0
acidic.
[H+], [OH-], and pH
[H+] × [OH-] = 1.0 × 10-14
[OH-] = (1.0 × 10-14)/ [H+]
= (1.0 × 10-14)/ 3.7×10-9
= 2.7 × 10-6 M
pH > 7.0
basic.
pH=-log[H+] = -log(3.7×10-9) = 8.43
Salt solution
Solutions of many salts are neutral in pH.
Both cation and anion can’t form weak base
or acid, the salt solution will be neutral in pH.
Solutions of many other salts are acidic or basic.
If the cation can
form weak base, the
salt solution will be
acidic.
If the anion can
form weak acid, the
salt solution will be
basic.
Salt solution
Question 2 – Determine if each salt will form a
solution that is acidic, basic, or pH-neutral.
a. Al(NO3)2
b. K2CO3
c. NaF
d. CaBr2
a. Cation Al3+ can form a weak base Al(OH)3, anion can’t
form a weak acid.
Hence, the solution of this salt will be acidic.
Salt solution
Question 2 – Determine if each salt will form a solution that is acidic, basic, or
pH-neutral.
a. Al(NO3)2
b. K2CO3
c. NaF
d. CaBr2
b. Cation K+ can’t form a weak base, anion CO32- can form
a weak acid.
Hence, the solution of this salt will be basic.
c. Cation Na+ can’t form a weak base, anion F- can form a
weak acid.
Hence, the solution of this salt will be basic.
d. ??????
Conjugate acid/base
conjugate
acid
HB (aq) ⇌ H+ (aq) +
base
B- (aq)
B- (aq) + H2O ⇌ HB (aq) + OH- (aq)
Ka of HB
Kb of B-
acid
base
conjugate
KaKb = Kw = 1.0 X 10-14
or
pKa + pKb = pKw = 14.00
Conjugate acid/base

HNO2 (aq)  H 2O(l )  NO2 (aq)  H 3O  (aq)

NO2 (aq)  H 2O(l )  HNO2 (aq)  OH  (aq)
Kw 1.0  1014
11
Kb 


2
.
2

10
Ka 4.6  104