Chem 162 2015 chapter 18 acid base solubility

Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 11

15.23d Calculate the pH of a mixture containing 0.100 M HC
3
H
5
O
2
and 0.100 M NaC
3
H
5
O
2
. (HC
3
H
5
O
2
, K
a
= 1.3 x 10
-5
)
*ET: Problem done previously. Note that there are fewer calculations when an uncharged substance reacts with H
2
O rather than a charged substance. Hence, I
chose to react HP with water rather than NaP. However, either is OK.
0.100 M HP
0.100 M NaP (Get rid of Na + spectator ion).
HP + H
2
O


H
3
O
+
+ P
-
HP(aq) +
H
2
O( l )


H
3
O
+
(aq) + P
-
(aq)
Initial 0.100 0 0.100
Change
Equilibrium
HP(aq) +
H
2
O( l )


H
3
O + (aq) + P - (aq)
Initial 0.100 0 0.100
Change -X +X +X
Equilibrium 0.100 - X +X 0.100+X
([H
3
O
+
][P
-
])/[HP] = K
a
((X) x (0.100 + X))/(0.100 - X) = 1.3 x 10 -5
Quadratic equation; simplify with small K rule:
(X x 0.100)/(0.100) = 1.3 x 10
-5
X = 1.3 x 10 -5
pH = -log 1.3 x 10
-5
= 4.89
Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 12

27d. Calculate the pH after 0.020 mol HCl is added to 1.00 L of a mixture containing
0.100 M HC
3
H
5
O
2
and 0.100 M NaC
3
H
5
O
2
. K
propanoic acid
= 1.3x10 -5
Note: In absence of a buffer, pH of 0.020M HCl = 1.70
ET: This slide ties into the previous slide of an HP/P
-
buffer without a strong acid.
ET: Given weak acid, conjugate base, and added strong acid concentrations; find pH.
What equilibrium equation should be used for the ICE table?
.
So
Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 13

HENDERSON-HASSELBALCH EQUATION
An equation for buffer solutions
(students not responsible for this derivation)
HA + H
2
O

 H
3
O + + A -
([H
3
O
+
][A
-
])/[HA] = K
a
[H
3
O + ] = K
a
x [HA]/[A - ]
-log[H
3
O
+
] = -logK
a
+ -log([HA]/[A
-
])
-log[H
3
O + ] = -logK
a
+ log([A - ]/[HA])
pH
3
O
+
= pK
a
+ log([A
-
]/[HA])
pH = pK
a
+ log([A - ]/[HA])
pH = pK
a