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CHXI07:EQUILIBRIUM

314906 The precipitate of $C a F_{2} (K_{S P} = 1 .7 \times 1 0^{- 10})$ is observed when equal volumes of the following are mixed.

1

1 0^{- 4} M C a^{2 +} + 1 0^{- 4} M F^{-}

2

1 0^{- 2} M C a^{2 +} + 1 0^{- 3} M F^{-}

3

1 0^{- 5} M C a^{2 +} + 1 0^{- 3} M F^{-}

4

1 0^{- 3} M C a^{2 +} + 1 0^{- 5} M F^{-}

Explanation:

The reaction is
${CaF}_{2} ⇌ {Ca}^{2 +} + 2 F^{-}$
$K_{S P} = [{Ca}^{2 +}] {[F^{-}]}^{2}$
(1) ${(K_{S P})}_{1} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 4}) {(0 .5 \times 1 0^{- 4})}^{2}$ $= 1.25 \times 10^{- 13}$
(2) ${(K_{SP})}_{2} = [C a^{2 +}] {[F^{-}]}^{2} = (0.5 \times 1 0^{- 2}) {(0.5 \times 1 0^{- 3})}^{2}$
$= 1 . 25 \times 1 0^{- 9}$
(3) ${(K_{S P})}_{3} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 5}) {(0 .5 \times 1 0^{- 3})}^{2}$
$= 1.25 \times 10^{- 12}$
(4) ${(K_{S P})}_{4} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 3}) {(0 .5 \times 1 0^{- 5})}^{2}$
$= 1.25 \times 10^{- 14}$

CHXI07:EQUILIBRIUM

314969 Read Assertion and Reason carefully to mark the correct option given below.
Assertion :
Addition of silver ions to a mixture of aqeous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl
Reason :
$K_{sp}$ of $AgCl < K_{sp}$ of AgBr

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXI07:EQUILIBRIUM

314907 The solubility product of chalk is $9.3 \times 10^{- 8}$ . Calculate its solubility in gram per litre

1 0.3040 gram/litre

2 0.0304 gram/litre

3 2.0304 gram/litre

4 4.0304 gram/litre

Explanation:

${CaCO}_{3} ⇌ {Ca}^{2 +} + {CO}_{3}^{2 -}$
Let the solubility of ${CaCO}_{3}$ be ' $s$ ' mole per litre
$∴ K_{sp} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = s . s$
$∴ s = \sqrt{K_{sp}} = \sqrt{9.3 \times 10^{- 8}} = 0.000304 mole / litre$
Solubility in $g / l =$ mole/litre $\times$ Molecular wt.
$= 0 . 000304 \times 100 = 0 .0304 gram/litre$

CHXI07:EQUILIBRIUM

314908 When equal volumes of ${AgNO}_{3}$ and $NaCl$ solution are mixed, the precipitation of $A g C l (K_{s p} = 1.81 \times 10^{- 10})$ will occur with

1

10^{- 3} M ({Ag}^{+})

and

10^{- 10} M ({Cl}^{-})

2

10^{- 5} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

3

10^{- 6} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

4

10^{- 4} M ({Ag}^{+})

and

10^{- 4} M ({Cl}^{-})

Explanation:

After mixing,
$\begin{array}{l} [A g^{+}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ [C l^{-}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ K_{s p} = [A g^{+}] [C l^{-}] = {(5 \times 10^{- 5})}^{2} \\ = 2.5 \times 10^{- 9} \end{array}$
Since, ionic product is greater than solubility product, hence precipitation will take place.

CHXI07:EQUILIBRIUM

314906 The precipitate of $C a F_{2} (K_{S P} = 1 .7 \times 1 0^{- 10})$ is observed when equal volumes of the following are mixed.

1

1 0^{- 4} M C a^{2 +} + 1 0^{- 4} M F^{-}

2

1 0^{- 2} M C a^{2 +} + 1 0^{- 3} M F^{-}

3

1 0^{- 5} M C a^{2 +} + 1 0^{- 3} M F^{-}

4

1 0^{- 3} M C a^{2 +} + 1 0^{- 5} M F^{-}

Explanation:

The reaction is
${CaF}_{2} ⇌ {Ca}^{2 +} + 2 F^{-}$
$K_{S P} = [{Ca}^{2 +}] {[F^{-}]}^{2}$
(1) ${(K_{S P})}_{1} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 4}) {(0 .5 \times 1 0^{- 4})}^{2}$ $= 1.25 \times 10^{- 13}$
(2) ${(K_{SP})}_{2} = [C a^{2 +}] {[F^{-}]}^{2} = (0.5 \times 1 0^{- 2}) {(0.5 \times 1 0^{- 3})}^{2}$
$= 1 . 25 \times 1 0^{- 9}$
(3) ${(K_{S P})}_{3} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 5}) {(0 .5 \times 1 0^{- 3})}^{2}$
$= 1.25 \times 10^{- 12}$
(4) ${(K_{S P})}_{4} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 3}) {(0 .5 \times 1 0^{- 5})}^{2}$
$= 1.25 \times 10^{- 14}$

CHXI07:EQUILIBRIUM

314969 Read Assertion and Reason carefully to mark the correct option given below.
Assertion :
Addition of silver ions to a mixture of aqeous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl
Reason :
$K_{sp}$ of $AgCl < K_{sp}$ of AgBr

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXI07:EQUILIBRIUM

314907 The solubility product of chalk is $9.3 \times 10^{- 8}$ . Calculate its solubility in gram per litre

1 0.3040 gram/litre

2 0.0304 gram/litre

3 2.0304 gram/litre

4 4.0304 gram/litre

Explanation:

${CaCO}_{3} ⇌ {Ca}^{2 +} + {CO}_{3}^{2 -}$
Let the solubility of ${CaCO}_{3}$ be ' $s$ ' mole per litre
$∴ K_{sp} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = s . s$
$∴ s = \sqrt{K_{sp}} = \sqrt{9.3 \times 10^{- 8}} = 0.000304 mole / litre$
Solubility in $g / l =$ mole/litre $\times$ Molecular wt.
$= 0 . 000304 \times 100 = 0 .0304 gram/litre$

CHXI07:EQUILIBRIUM

314908 When equal volumes of ${AgNO}_{3}$ and $NaCl$ solution are mixed, the precipitation of $A g C l (K_{s p} = 1.81 \times 10^{- 10})$ will occur with

1

10^{- 3} M ({Ag}^{+})

and

10^{- 10} M ({Cl}^{-})

2

10^{- 5} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

3

10^{- 6} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

4

10^{- 4} M ({Ag}^{+})

and

10^{- 4} M ({Cl}^{-})

Explanation:

After mixing,
$\begin{array}{l} [A g^{+}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ [C l^{-}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ K_{s p} = [A g^{+}] [C l^{-}] = {(5 \times 10^{- 5})}^{2} \\ = 2.5 \times 10^{- 9} \end{array}$
Since, ionic product is greater than solubility product, hence precipitation will take place.

CHXI07:EQUILIBRIUM

314906 The precipitate of $C a F_{2} (K_{S P} = 1 .7 \times 1 0^{- 10})$ is observed when equal volumes of the following are mixed.

1

1 0^{- 4} M C a^{2 +} + 1 0^{- 4} M F^{-}

2

1 0^{- 2} M C a^{2 +} + 1 0^{- 3} M F^{-}

3

1 0^{- 5} M C a^{2 +} + 1 0^{- 3} M F^{-}

4

1 0^{- 3} M C a^{2 +} + 1 0^{- 5} M F^{-}

Explanation:

The reaction is
${CaF}_{2} ⇌ {Ca}^{2 +} + 2 F^{-}$
$K_{S P} = [{Ca}^{2 +}] {[F^{-}]}^{2}$
(1) ${(K_{S P})}_{1} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 4}) {(0 .5 \times 1 0^{- 4})}^{2}$ $= 1.25 \times 10^{- 13}$
(2) ${(K_{SP})}_{2} = [C a^{2 +}] {[F^{-}]}^{2} = (0.5 \times 1 0^{- 2}) {(0.5 \times 1 0^{- 3})}^{2}$
$= 1 . 25 \times 1 0^{- 9}$
(3) ${(K_{S P})}_{3} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 5}) {(0 .5 \times 1 0^{- 3})}^{2}$
$= 1.25 \times 10^{- 12}$
(4) ${(K_{S P})}_{4} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 3}) {(0 .5 \times 1 0^{- 5})}^{2}$
$= 1.25 \times 10^{- 14}$

CHXI07:EQUILIBRIUM

314969 Read Assertion and Reason carefully to mark the correct option given below.
Assertion :
Addition of silver ions to a mixture of aqeous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl
Reason :
$K_{sp}$ of $AgCl < K_{sp}$ of AgBr

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXI07:EQUILIBRIUM

314907 The solubility product of chalk is $9.3 \times 10^{- 8}$ . Calculate its solubility in gram per litre

1 0.3040 gram/litre

2 0.0304 gram/litre

3 2.0304 gram/litre

4 4.0304 gram/litre

Explanation:

${CaCO}_{3} ⇌ {Ca}^{2 +} + {CO}_{3}^{2 -}$
Let the solubility of ${CaCO}_{3}$ be ' $s$ ' mole per litre
$∴ K_{sp} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = s . s$
$∴ s = \sqrt{K_{sp}} = \sqrt{9.3 \times 10^{- 8}} = 0.000304 mole / litre$
Solubility in $g / l =$ mole/litre $\times$ Molecular wt.
$= 0 . 000304 \times 100 = 0 .0304 gram/litre$

CHXI07:EQUILIBRIUM

314908 When equal volumes of ${AgNO}_{3}$ and $NaCl$ solution are mixed, the precipitation of $A g C l (K_{s p} = 1.81 \times 10^{- 10})$ will occur with

1

10^{- 3} M ({Ag}^{+})

and

10^{- 10} M ({Cl}^{-})

2

10^{- 5} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

3

10^{- 6} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

4

10^{- 4} M ({Ag}^{+})

and

10^{- 4} M ({Cl}^{-})

Explanation:

After mixing,
$\begin{array}{l} [A g^{+}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ [C l^{-}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ K_{s p} = [A g^{+}] [C l^{-}] = {(5 \times 10^{- 5})}^{2} \\ = 2.5 \times 10^{- 9} \end{array}$
Since, ionic product is greater than solubility product, hence precipitation will take place.

CHXI07:EQUILIBRIUM

314906 The precipitate of $C a F_{2} (K_{S P} = 1 .7 \times 1 0^{- 10})$ is observed when equal volumes of the following are mixed.

1

1 0^{- 4} M C a^{2 +} + 1 0^{- 4} M F^{-}

2

1 0^{- 2} M C a^{2 +} + 1 0^{- 3} M F^{-}

3

1 0^{- 5} M C a^{2 +} + 1 0^{- 3} M F^{-}

4

1 0^{- 3} M C a^{2 +} + 1 0^{- 5} M F^{-}

Explanation:

The reaction is
${CaF}_{2} ⇌ {Ca}^{2 +} + 2 F^{-}$
$K_{S P} = [{Ca}^{2 +}] {[F^{-}]}^{2}$
(1) ${(K_{S P})}_{1} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 4}) {(0 .5 \times 1 0^{- 4})}^{2}$ $= 1.25 \times 10^{- 13}$
(2) ${(K_{SP})}_{2} = [C a^{2 +}] {[F^{-}]}^{2} = (0.5 \times 1 0^{- 2}) {(0.5 \times 1 0^{- 3})}^{2}$
$= 1 . 25 \times 1 0^{- 9}$
(3) ${(K_{S P})}_{3} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 5}) {(0 .5 \times 1 0^{- 3})}^{2}$
$= 1.25 \times 10^{- 12}$
(4) ${(K_{S P})}_{4} = [C a^{2 +}] {[F^{-}]}^{2} = (0 .5 \times 1 0^{- 3}) {(0 .5 \times 1 0^{- 5})}^{2}$
$= 1.25 \times 10^{- 14}$

CHXI07:EQUILIBRIUM

314969 Read Assertion and Reason carefully to mark the correct option given below.
Assertion :
Addition of silver ions to a mixture of aqeous sodium chloride and sodium bromide solution will first precipitate AgBr rather than AgCl
Reason :
$K_{sp}$ of $AgCl < K_{sp}$ of AgBr

1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.

2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.

3 Assertion is correct but Reason is incorrect.

4 Assertion is incorrect but reason is correct.

CHXI07:EQUILIBRIUM

314907 The solubility product of chalk is $9.3 \times 10^{- 8}$ . Calculate its solubility in gram per litre

1 0.3040 gram/litre

2 0.0304 gram/litre

3 2.0304 gram/litre

4 4.0304 gram/litre

Explanation:

${CaCO}_{3} ⇌ {Ca}^{2 +} + {CO}_{3}^{2 -}$
Let the solubility of ${CaCO}_{3}$ be ' $s$ ' mole per litre
$∴ K_{sp} = [{Ca}^{2 +}] [{CO}_{3}^{2 -}] = s . s$
$∴ s = \sqrt{K_{sp}} = \sqrt{9.3 \times 10^{- 8}} = 0.000304 mole / litre$
Solubility in $g / l =$ mole/litre $\times$ Molecular wt.
$= 0 . 000304 \times 100 = 0 .0304 gram/litre$

CHXI07:EQUILIBRIUM

314908 When equal volumes of ${AgNO}_{3}$ and $NaCl$ solution are mixed, the precipitation of $A g C l (K_{s p} = 1.81 \times 10^{- 10})$ will occur with

1

10^{- 3} M ({Ag}^{+})

and

10^{- 10} M ({Cl}^{-})

2

10^{- 5} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

3

10^{- 6} M ({Ag}^{+})

and

10^{- 5} M ({Cl}^{-})

4

10^{- 4} M ({Ag}^{+})

and

10^{- 4} M ({Cl}^{-})

Explanation:

After mixing,
$\begin{array}{l} [A g^{+}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ [C l^{-}] = (1 / 2) \times 10^{- 4} = 5 \times 10^{- 5} M \\ K_{s p} = [A g^{+}] [C l^{-}] = {(5 \times 10^{- 5})}^{2} \\ = 2.5 \times 10^{- 9} \end{array}$
Since, ionic product is greater than solubility product, hence precipitation will take place.

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