Buffers And Titrations Worksheet With Answer Key Page 25
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26-
-
The conjugate base OCl
is more stable than OI
because Cl, being more
electronegative, is better able to accommodate the negative charge.
O ↔ OCl
-
+
(b) HOCl + H
+ H
O
2
3
OR
HOCL ↔ OCl
-
+
+ H
-
(ci) K
= [HOCl][OH
]
b
-
[OCl
]
-14
-8
-7
(ii) K
= K
/K
=(1.0x10
/2.9x10
)=3.4x10
b
w
a
(iii) K = (x)(x)/(1.2-x) ≈ x
2
/1.2
2
-7
x
= (1.2)( 3.4x10
)
-
-4
x = [OH
] = 6.4 x 10
M
+
-6.48
-7
(di) [H
] = 10
= 3.3 x 10
M
+
-
-8
(ii) K
= [H
][OCl
] = 2.9 x 10
a
[HOCl]
= 0.088 [HOCl] > [OCl
-
-8
-
[OCl
] = 2.9 x 10
]
-7
[HOCl]
3.3 x 10
6)
(a) n
= 14.84 g C
H
NH
/ 45.09 g/mol C
H
NH
= 0.3293 mol C
H
NH
C2H5NH2
2
5
2
2
5
2
2
5
2
M = 0.3293 mol C
H
NH
/ 5.00 L = 0.659 M
2
5
2
+
-
(b) K
= [C
H
NH
][OH
]
b
2
5
3
[C
H
NH
]
2
5
2
(c) C
H
NH
is present in the solution at the higher concentration at equilibrium.
2
5
2
Ethylamine is a weak base, and thus is has a small K
value. Therefore, only partial
b
+
dissociation of C
H
NH
occurs in water, and [C
H
NH
] is thus less than [C
H
NH
].
2
5
2
2
5
3
2
5
2
+
(di) pH = -log[H
]
+
-10.93
-11
[H
] = 10
= 1.17 x 10
-
+
-4
[OH
] = K
/ [H
] = 8.5 x 10
M
w
OR
pOH = 14 – pH = 3.07
-
-3.07
-4
[OH
] = 10
= 8.5 x 10
↔ C
+
+
(ii) C
H
NH
+ H
O
H
NH
+ H
O
2
5
2
3
2
5
3
2
(iii) moles of C
H
NH
= (0.5 x 0.5 mol / 1 L) = 0.250 mol
2
5
2
+
Moles of H
O
= (.5 x .2 mol/ 1 L) = 0.100 mol
3
+
+
[C
H
NH
]
[H
O
]
[C
H
NH
]
2
5
2
3
2
5
3
Initial
0.250
0.100
~0
Change
-0.100
-0.100
+0.100
Final Value
0.150
~0
0.100
(iv) [C
H
NH
] = 0.150 M
2
5
2
+
-
-4
-4
K
= [C
H
NH
][OH
] = (0.1)(8.5 x 10
) = 5.67 x 10
b
2
5
3
[C
H
NH
]
0.150
2
5
2
7)
+
-
a) K
= [H
][C
H
O
] ; 0.50 m X 0.0166 = 0.0083 M = x
a
3
5
3
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