MHS  AP Chemistry
Name _______________________________
Period ___ Date ___/___/___
17  Acid-Base Reactions
1.
COMMON ION
Circle pairs of molecules that fit the “common ion” definition:
HF & KF
HF
HCH3COO
HCN
2.
HCH3COO & NaCl
Ka = 6.6 x 10-4
Ka = 1.8 x 10-5
Ka = 6.2 x 10-10
NH3 & NH4Cl
ammonia, NH3
methylamine, CH3NH2
aniline, C6H5NH2
EFFECT
H2S & NaOH
Kb = 1.8 x 10-5
Kb = 4.4 x 10-4
Kb = 3.8 x 10-10
For the molecules you did not circle in #1, identify a common ion.
HCH3COO: LiCH3COO
NaCl: HCl
H2S: Na2S
NaOH: Ca(OH)2
3.
Consider a mixture of HF and KF.
a) Write the acid dissociation equation for HF.
HF  H+ + F
This is the same equation used for the common ion effect.
b)
I
C
E
Calculate the pH of a 0.200 M solution of HF.
HF

0.200
-x
~0.200
H+
0
+x
x
F0
+x
x
+
Ka = x2/0.200 = 6.6 x 10-4
X = [H+] = 0.0115 M
pH = - log (0.0115 M ) = 1.940
** assume x is very small, so (0.200-x) = 0.200
c)
Using Le Chatelier’s Principle, predict how the concentrations of reactants and products
will change when KF is added to the HF solution.
HF(aq) + H2O(l)  H3O+(aq) + F(aq)
add F

X


When KF is added to an HF solution, the pH of the solution will increase (become more
basic).
d)
Use an ICE box to set up and calculate the pH of a solution that is initially 0.200 M HF
and 0.100 M F.
HF(aq) + H2O(l)  H3O+(aq) + F(aq)
I
0.200
0
0.100
C
-x
+x
+x
E
~0.200
x
~0.100
Ka = (0.100)(x)/0.200 = 6.6 x 10-4
X = [H+] = 0.0132 M
pH = - log (0.0132 M ) = 2.879
** assume x is very small
e)
Compare the pH you calculated in b) and the pH you calculated in d). Did this change match the shift
you predicted in c)?
The pH in b) 1.940 was and the pH in d) was 2.879. In c) we predicted
that the pH would increase. The calculations match our predictions.
Practice Problems: On the back side of this paper (and additional paper if necessary) solve these
problems from your text book.
Practice Exercise 17.1 (p. 721), Practice Exercise 17.2 (p. 722), problems 17.13, 17.15, and 17.17 (p. 759)
Answers in book.
Practice Problems:
A solution is made by adding solid NaCN to a 0.0300 M HCN solution until the [NaCN] = 0.0150 M.
Calculate the pH of the solution when it reaches equilibrium.
3.
I
C
E
HCN
 H+
0.0300
0
-x
+x
~0.0300
x
+
CN0.0150
+x
~0.0150
Ka = (0.0150)(x)/0.0300 = 6.2 x 10-10
X = [H+] = 1.24 x 10-9 M
pH = - log (1.24 x 10-9 M) = 8.907
** assume x is very small
A solution initially has [NH3] = 0.100 M and [NH4+] = 0.0500 M. Calculate the pH of the solution
after it reaches equilibrium.
4.
NH3(aq) + H2O(l)  OH + NH4+
Kb = (0.0500)(x)/0.100 = 1.8 x 10-5
I
0.100
0
0.0500
C
-x
+x
+x
E
~0.100
x
~0.0500
pH = 14-4.444 = 9.556
** assume x is very small
5.
X = [OH] = 3.60 x 10-5 M
pOH = - log (3.60 x 10-5M ) = 4.444
A mixture is formed by combining 200. mL of 0.200 M HC2H3O2 with 300. mL of 0.0500 M
KC2H3O2.
(a) What are the initial concentrations of HC2H3O2 and C2H3O2?
Use M1V1 = M2V2 to determine the final concentrations.
Final volume, V2, is 200. mL + 300. mL = 500. mL
[HC2H3O2]: (0.200 M) x (200. mL) = M2 x (500. mL)
-
[C2H3O2 ]:
M2 = 0.0800 M
(0.0500 M) x (300. mL) = M2 x (500. mL) M2 = 0.0300 M
(b) What is the pH of the equilibrium mixture?
HC2H3O2
I
C
E
0.0800
-x
~0.0800
** assume x is very small

H+
0
+x
x
+
C2H3O2
-
0.0300
+x
~0.0300
Ka = (0.0300)(x)/0.0800 = 1.8 x 10-5
X = [H+] = 4.80 x 10-5 M
pH = - log (4.80 x 10-5 M) = 4.319