2M solution of Na2CO3 is boiled in a closed container with excess of CaF2. Very little amount of CaCO3 and NaF are formed. If the solubility product (Ksp) of CaCO3 is x and molar solubility of CaF2 is y, find the molar concentration of F– in resulting solution after equilibrium is attained.
Answers
Answer:
Na
2
CO
3
+CaF
2
(s)⟶CaCO
3
+2NaF
Mole taken 2 0 0
Mole left (2-a) a 2a
where a is very-very small and thus assumes that CaCO
3
is an insoluble form
Now, K
sp
of CaCO
3
=x=[Ca
2+
][CO
3
2−
]
Also [CO
3
2−
]=2−a+a=2∴[Ca
2+
]=
2
x
For CaF
2
⇌Ca
2+
+2F
⊝
K
sp(CaF
2
)
=[Ca
2+
][F
⊝
]
2
=(y)(2y)
2
=4y
3
.
Further for [F
⊝
], we can have
[F
⊝
]=[F
⊝
] from CaF
2
+[F
⊝
] from NaF
[F
⊝
]=
[Ca
2+
]
K
sp(CaF
2
)
+Negligible value
[F
⊝
]=
x/2
4y
3
=
x
8y
3
Explanation:
Answer
Na
2
CO
3
+CaF
2
(s)⟶CaCO
3
+2NaF
Mole taken 2 0 0
Mole left (2-a) a 2a
where a is very-very small and thus assumes that CaCO
3
is an insoluble form
Now, K
sp
of CaCO
3
=x=[Ca
2+
][CO
3
2−
]
Also [CO
3
2−
]=2−a+a=2∴[Ca
2+
]=
2
x
For CaF
2
⇌Ca
2+
+2F
⊝
K
sp(CaF
2
)
=[Ca
2+
][F
⊝
]
2
=(y)(2y)
2
=4y
3
.
Further for [F
⊝
], we can have
[F
⊝
]=[F
⊝
] from CaF
2
+[F
⊝
] from NaF
[F
⊝
]=
[Ca
2+
]
K
sp(CaF
2
)
+Negligible value
[F
⊝
]=
x/2
4y
3
=
x
8y
3