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

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
$\begin{matrix} C a F_{2} (s) & ⇌ & C a^{2 +} & + & 2 F^{-} s & 2 s \end{matrix}$
$4 s^{3} = K_{s p} = 4 \times 10^{- 11} \Rightarrow s = 2.15 \times 10^{- 4} M$
So, amount of $F^{-}$ in $20000 L$ of water $= 2 s \times 20000 = 8.6 mol$

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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)

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}

)

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 tank contains

40 l

of solution containing

2 g

of substance per liter. Salt water containing

3 g

of this substance per liter runs in at the rate of

4 l / m i n

and the well-stirred mixture runs out at the same rate. Find the amount of substance in the tank after

15 m i n u t e s

(in grams).

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.

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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(101/3=2.15)

The correct option is C 8.6 molCaF2(s)⇌Ca2++2F−s2s 4s3=Ksp=4×10−11⇒s=2.15×10−4 M So, amount of F− in 20000 L of water =2s×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? $K_{s p} = 4 \times 10^{- 11} (10^{1 / 3} = 2.15)$

The correct option is C 8.6 mol
$\begin{matrix} C a F_{2} (s) & ⇌ & C a^{2 +} & + & 2 F^{-} s & 2 s \end{matrix}$
$4 s^{3} = K_{s p} = 4 \times 10^{- 11} \Rightarrow s = 2.15 \times 10^{- 4} M$
So, amount of $F^{-}$ in $20000 L$ of water $= 2 s \times 20000 = 8.6 mol$