Ap Solubility Set I Worksheet Page 3
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1
2
3=
+
−
2
2
K
[
Fe
][
OH
]
(b)
sp
+2
-3
-5
[Fe
] = (1.43x10
g Fe(OH)
/L)(1 mol Fe(OH)
/91.8617 g Fe(OH)
) = 1.55 x 10
M
2
2
2
=
=
-5
-5
2
-14
K
[
1.55
x
10
][
( 2
1.55
x
10
] )
1.50 x 10
sp
-
-5
-5
(c) pH means you are solving for [OH
], but you’ve already been given that (2*1.55x10
) = 3.11 x 10
M
-5
pH = 14-pOH = 14 – (-log (3.11 x 10
)) = 9.49
(d) Use the Ksp expression to calculate Q:
+2
-3
2+
[Fe
]
= 3.00 x 10
M (50mL Fe
/100mL solution) = .0015M
-
-6
2+
-6
[OH
]
= 4.00 x 10
M (50mL Fe
/100mL solution) = 2 x 10
M
2+
-
2
-6
2
-15
Q
= [Fe
][OH
]
= (.0015)(2 x 10
)
= 6 x 10
Q
< K
, so no ppt forms.
sp
sp
sp
2+
-
2
-3
-3
2
5. (a) K
= [Mg
][F
]
= (1.21×10
)(2.42×10
)
sp
-9
= 7.09×10
2+
(b) x = concentration of Mg
ion (by equilibrium)
-
2x = concentration of F
ion (by equilibrium)
2+
[Mg
] = x M
-
[F
] = (0.100 + 2x) M
since X is a small number then (0.100 + 2x) ≈ 0.100
-9
2
K
= 7.09×10
= (x)(0.100)
sp
-7
X = 7.09×10
2+
-3
-3
-7
[Mg
]= 1.21×10
-1.20929×10
= 7.09×10
M
M × 100.0 mL/300.0 mL = 1.00×10
2+
-3
-3
(c)
[Mg
] = 3.00×10
M
M × 200.0 mL/300.0 mL = 1.33×10
-
-3
-3
[F
] = 2.00×10
M
-3
-3
2
-9
trial K
= (1.00×10
)(1.33×10
)
= 1.78×10
sp
, ∴ no ppt.
-9
trial K
< = 7.09×10
sp
-3
(d) @ 18ºC, 1.21×10
M MgF
dissolves
2
-3
@ 27ºC, 1.17×10
M MgF
dissolves
2
↔ Mg
2+
-
MgF
+ 2 F
+ heat
2
dissolving is exothermic; if heat is increased it forces the equilibrium to shift left (according to LeChatelier’s
Principle) and less MgF
will dissolve.
2
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