Ap Exam Prep Stoichiometry Page 4

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+
+
pH = –log[H
]; 2.22 = –log[H
]
[H
] = M = [Asa
]
+
[HAsa] = 0.133 M – 6.03  10
-3
M = 0.127 M
(6.03  10
+
-3
2
[H
][Asa
]
)
-4
K =
=
= 2.85 10
[HAsa]
0.127
OR
when the solution is half-neutralized, pH = pK
a
–pH
at 10.00 mL, pH = 3.44; K = 10
–3.44
-4
= 10
= 3.6310
(ii) 0.025 L  0.100 mol/L = 2.50  10
-3
mol OH
-
2.50  10
- 2.00  10
mol neutralized = 5.0  10
-3
mol OH
-3
-4
mol OH
remaining in (25 + 15 mL) of
-
-
-
-4
solution; [OH
] = 5.010
mol/0.040 L = 0.0125 M
-
pH = 14 – pOH = 14 + log[OH
] = 14 – 1.9 = 12.1
2000 B:
total mass of carbon
24.022
 100 =
 100 = 15.9%
(a)
molar mass
151.03
1 mol hyd.
1 mol anhyd.
97.03 g anhyd.
(b) (i) 3.21 g 
= 2.06 g
151.03 g
1 mol hyd.
1 mol anhyd.
3.21 g – 2.06 g
(ii) mol H
O =
= 0.06375 mol
2
18.0 g/mol
L atm
0.06375 mol 0.0821
220 273 K
nRT
mol K
V =
=
= 2.67 L H
O
(g)
2
P
735
atm
760
2–
(c) (i) C
O
(aq)
2
4
-
0.0150 mol MnO
4
(ii) 17.80 mL 
-
–4
= 2.6710
mol MnO
4
1000 mL
2-
5 mol C
O
2
4
-
2-
2.6710
–4
mol MnO
= 6.6810
–4
mol C
O
4
2
4
-
2 mol MnO
4
2-
6.6810
–4
mol C
O
2
4
(iii) 100. mL 
2-
= 3.3410
–3
mol C
O
2
4
20 mL
1 mol BeC
O
97.03 g
2
4
2-
–3
(iv) 3.3410
mol C
O
= 0.324 g BeC
O
(s)
2
4
2
4
2-
1 mol
1 mol C
O
2
4
0.324 g BeC
O
(s)
2
4
 100 = 93.9%
0.345 g sample

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