Equilibrium III: Basic Acid-Base Equilibrium
1. What are the species (equilibrium) concentrations of H3O+, OH-, and Cl- in 0.10 M hydrochloric
acid?
[H3 O + ] = [Cl− ] = 0.10 M
[OH − ] =
pH = − log(0.10) = 1.0
Kw
1.0 × 10
=
+
0.10
[H3 O ]
−14
= 1.0 × 10−13 M
2. What are the concentrations of each species in a solution prepared to be 0.10 M acetic acid
(CH3COOH)? Ka = 1.75 × 10-5
HOAc
+
H2O
OAc-
R
at equilibrium 0.10 − x
Ka =
−
x
+
+
H3 O+
x
2
[OAc ][H 3 O ]
x
= 1.75 × 10−5 =
x may be small compared to 0.10, so...
[HOAc]
0.10 − x
1.75 × 10−5 ≅
x2
0.10
x = [OAc − ] = [H 3 O + ] = 1.75 × 10−6 = 1.33 × 10−3 M
Check to see if assumption is acceptable:
0.00133
× 100 = 1.3% (< 5%, so accept assumption)
0.1
[HOAc] = 0.10 M − 0.00133 M = 0.099 M
[OH - ] =
1.0 × 10−14
= 7.6 × 10−12 M
0.00133
3. What is the approximate pH of 0.075 M formic acid? Ka = 1.7 × 10-4
This problem is virtually identical to problem 2.
4. What is the Ka of nitrous acid if a 0.050 M solution has a pH of 2.34?
[H 3 O + ] = 10−2.34 = 4.57 × 10−3 M ( = [NO −2 ])
CHNO2 = 0.050 M [HNO 2 ] = 0.050 M − 4.57 × 10−3 M = 0.0454 M
Ka =
(4.57 × 10−3 ) 2
= 4.6 × 10−4
0.0454
5. What is the pH of 0.010 M ammonia? Kb = 1.8 × 10-5
NH 3
+
H2 O
R
NH +4
at equilibrium 0.010 − x
Ka =
+
OH −
x
x
[NH 4+ ][OH − ]
x2
= 1.8 × 10−5 =
x is small compared to 0.010, so...
[NH 3 ]
0.010 − x
x2
x = [NH 4+ ] = [OH − ] = 1.8 × 10−7 = 4.24 × 10−4 M
0.010
(Demonstrate yourself that x is small compared to 0.010 M)
1.8 × 10−5 ≅
[NH 3 ] = 0.010 M − 4.24 × 10−4 M = 9.6 × 10−3 M
[H 3 O + ] =
1.0 × 10−14
= 2.36 × 10−11 M
−4
4.24 × 10
pH = − log(2.36 × 10 −11 ) = 10.6
6. Assuming that Ka = 1.75 × 10-5 for acetic acid, what is Kb for acetate ion? What is the pH of
0.010 M sodium acetate?
Kb =
K w 1.0 × 10−14
=
= 5.71 × 10−10
K a 1.75 × 10−5
OAc −
+
R
H2 O
HOAc
at equilibrium 0.010 − x
5.71 × 10−10 =
+
x
−
2
2
[HOAc][OH ]
x
x
=
≅
−
0.010 − x 0.010
[OAc ]
x = [HOAc] = [OH − ] = 2.4 × 10−6 M
pH = 14 − pOH = 8.38
pOH = 5.62
OH −
x