AP CHEM WKST KEY: Sem 2 Review Part 3
1) 2AgNO3 + Na2CrO4 → Ag2CrO4(s) + 2NaNO3
250.00 mL 250.00 mL
0.175 M
0.250 M
- find [reactants] after mixing using M1V1 = M2V2:
[AgNO3] =
(0.175 M)(250 .00 mL)
= 0.0875 M AgNO3
500 .00 mL
[Na2CrO4] =
(0.250 M)(250 .00 mL)
= 0.125 M Na2CrO4
500 .00 mL
- determine if precipitate forms:
Ag2CrO4(s) ⇄ 2Ag+ + CrO42−
0.0875 M 0.125 M
Q = [Ag+]2[CrO42−] = (0.0875)2(0.125) = 9.57 x 10−4
Since Q > K, there will be a precipitate YES
- finding [ions]:

- find limiting reactant first: 0.0875 M AgNO3  1 mol Na 2 Cr O4
 2 mol AgNO
3

 AgNO3 is L.R.

 = 0.0438 M Na2CrO4


- ice table:
[ ]i
0.0875
0.0875
0.250
0.125
Ag+
NO3−
Na+
CrO42−
Δ[ ]
−0.0875
0
0
−0.0483
[ ]e
−−
0.0875
0.250
0.081
[Ag+] =
[NO3−] =
[Na+] =
[CrO42−] =
 1 mol Cr O4 2 
- find amount of CrO42− to react with Ag+: 0.0875 M Ag+ 
 2 mol Ag 

- use Ksp to determine the [Ag+]:
Ag2CrO4(s) ⇄ 2Ag+ + CrO42−
2x
0.081 + x
1.1 x 10−5 M
0.0875 M
0.250 M
0.081 M

 = 0.0438 M CrO42−


Ksp = 9.0 x 10−12
Ksp = [Ag+]2[CrO42−]
9.0 x 10−12 = (2x)2(0.081)
9.0 x 10−12 = 0.32x2
2.8 x 10−11 = x2
x = 5.3 x 10−6
2) Ba3(PO4)2(s) ⇄ 3Ba2+ + 2PO43−
3x
3
2x
Ksp = 6 x 10−39
2
6 x 10-39 =[Ba2+ ] [PO34 ]
6 x 10−39 = (3x)3(2x)2 = 108 x5
6 x10−41 = x5
x = 9 x 10−9 M
______________________________________________________________________________________________________
3) a) acidic
b) acidic
c) acidic
d) acidic
e) neutral
f) basic
g) acidic
4)
ACIDS
Al(H2O)63+
Cu(H2O)42+
BASES
Al(H2O)5(OH)2+
Cu(H2O)3(OH)+
5) acid → AlCl3
base → NH3
pH = −log[H+] = −log(0.0326 M) = 1.49
6) a) [HNO3] = 0.0326 M
HF ⇄ H+ + F−
b) [HF] = 0.0226 M
0.0226-x
Ka 
x
Ka = 7.2 x 10−4
x
[H ][F  ]
[HF ]
(x)(x)
0.0226  x
1.6 x 10−5 − 7.2 x 10−4x = x2
0 = x2 + 7.2 x 10−4x − 1.6 x 10−5
x = 0.0037, −0.0044
[H+] = 0.0037
7.2 x 10−4 =
NH3 + H2O ⇄ NH4+ + OH−
c) [NH3] = 1.50 M
1.50-x
x
H3PO4 ⇄ H+ + H2PO4−
d) [H3PO4] = 1.50 M
1.50-x
K a1 
x
Kb = 1.8 x 10−5
x
[NH 4 ][OH ]
Kb 
[NH 3 ]
x
[H ][H2PO4 ]
[H3PO4 ]
pH = 2.43
(x)(x)
1.50
2.7 x 10−5 = x2
x = 0.0052 M = [OH−]
1.8 x 10−5 =
pOH = 2.28
pH = 11.72
Ka1 = 7.5 x 10−3
(x)(x)
1.50  x
0.011 − 0.0075x = x2
0 = x2 + 0.0075x − 0.011
x = 0.10, −0.11
[H+] = 0.10 M
7.5 x 10−3 =
pH = 1.00
7) 50.0 mL 1.25 M HNO3/ 0.250 M NaOH
a) 10.0 mL NaOH is added
mmoli
mmolf
H+
62.5
60.0
+
OH−
2.50
0
→
H2O
[H+ ] =
60.0 mmol
= 1.00 M
60.0 mL
pH = -log(1.00) = 0
b) 350.0 mL NaOH added
mmoli
mmolf
H+
62.5
0
+
OH−
87.5
25.0
→
H2O
[OH- ] =
25.0 mmol
= 0.0625 M
400.0 mL
pOH = -log(0.0625) = 1.204
pH = 14 – pOH = 14 – 1.204 = 12.796
____________________________________________________________________________________________
8) 0.125 M HC7H5O2 / 0.150 M NaC7H5O2
Ka = 6.4 x 10−5
 [C H O  ] 
a) pH = pKa + log  7 5 2 
 [HC H O ] 
7 5 2 

 0.150 M 
 = 4.19 + 0.0792 = 4.27
pH = −log(6.4 x 10−5) + log 

 0.125 M 
b) 0.0300 mol KOH added to 2 L of buffer
HC7H5O2 + OH− → H2O + C7H5O2−
HC7H5O2
OH−
C7H5O2−
[ ]i
0.125
0.0150
0.150
[ ]f
0.110
0
0.165
 [C H O  ] 
pH = pKa + log  7 5 2 
 [HC H O ] 
7 5 2 

 0.165 M 
 = 4.19 + 0.176 = 4.37
pH = −log(6.4 x 10−5) + log 

 0.110 M 