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Standard XII

Chemistry

Solubility Product

A well is dug...

Question

A well is dug in a bed of rock containing fluorspar $(C a F_{2})$ . If the well contains 20000 L of water, what is the amount of $F^{-}$ in it? ( $K_{s p} = 4 \times 10^{- 11}$ )

4.3 mol

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6.8 mol

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8.6 mol

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13.6 mol

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Solution

The correct option is C 8.6 mol
The dissociation equation is $C a F_{2} (s) ⇌ C a^{2 +} (a q) + 2 F^{-} (a q) t = 0100 a t e q b 1 - s s 2 s$

$K_{s p} = [C a^{2 +}] + [F^{-}]^{2}$
$K_{s p} = s \times (2 s)^{2} = 4 s^{3}$
$K_{s p} = 4 s^{3} = 4 \times 10^{- 11}$
$\Rightarrow s = 2.15 \times 10^{- 4} M$
So, amount of $F^{-}$ in 20000 L of water $= 2 s \times 20000 = 2 \times 2.15 \times 10^{- 4} \times 20000 = 8.6 m o l$

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Similar questions

Q. A well is dug in a bed of rock containing fluorspar

(C a F_{2})

. If the well contains

20000 L

of water, what is the amount of

F^{-}

in it?

K_{s p} = 4 \times 10^{- 11} (10^{1 / 3} = 2.15)

A well is dug in a bed of rock containing fluorspar

(C a F_{2})

. If the well contains

20000

L of water, what is the amount of

F^{-}

in it ? (

K_{s p} = 4 \times 10^{- 11}

)

3 M

solution of

N a_{2} C O_{3}

is boiled in a closed container with excess of

C a F_{2} .

Very little amount of

C a C O_{3}

and

N a F

are formed. If the solubility product

K_{s p}

C a C O_{3}

is y and molar solubility of

C a F_{2}

in water is x, find the molar concentration of

F^{-}

in resulting solution after equilibrium is attained.

Q. A solution of saturated

C a F_{2}

is found to contain

4.1 \times 10^{- 4} M

fluoride ion. Calculate the

K_{s p}

C a F_{2}

. Neglect hydrolysis.

3 M

solution of

N a_{2} C O_{3}

is boiled in a closed container with excess of

C a F_{2} .

Very little amount of

C a C O_{3}

and

N a F

are formed. If the solubility product

K_{s p}

C a C O_{3}

is y and molar solubility of

C a F_{2}

in water is x, find the molar concentration of

F^{-}

in resulting solution after equilibrium is attained.

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Solubility Product

Standard XII Chemistry

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A well is dug in a bed of rock containing fluorspar (CaF2). If the well contains 20000 L of water, what is the amount of F− in it? (Ksp=4×10−11)

The correct option is C 8.6 molThe dissociation equation is CaF2(s)⇌Ca2+(aq)+2F−(aq)t=0 1 0 0at eqb 1−s s 2s Ksp=[Ca2+]+[F−]2 Ksp=s×(2s)2=4s3 Ksp=4s3=4×10−11 ⇒s=2.15×10−4M So, amount of F− in 20000 L of water =2s×20000=2×2.15×10−4×20000=8.6 mol

A well is dug in a bed of rock containing fluorspar $(C a F_{2})$ . If the well contains 20000 L of water, what is the amount of $F^{-}$ in it? ( $K_{s p} = 4 \times 10^{- 11}$ )