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11th
Chemistry
Equilibrium
$Solubility Equilibria$

Solubility Equilibria

Equilibrium

Chemistry

Class 11

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Derivation and Problems on Solubility Products

15 mins

Applications of Solubility Product and Common-Ion Effect

18 mins

Problems on Common-Ion Effect in Solubility

15 mins

Study Materials

Law of Chemical Equilibrium, Equilibrium Constant - Examples

Revise with Concepts

Solubility and Solubility Product of sparingly soluble salts

ExampleDefinitionsFormulaes

Common Ion Effect on Solubility

ExampleDefinitionsFormulaes

What is the ratio of moles of $M g (O H)_{2}$ and $A l (O H)_{3}$ present in 1 lit saturated aqueous solution of $M g (O H)_{2}$ & $A l (O H)_{3} (K_{s p} o f M g (O H)_{2} = 4 \times 1 0^{- 12}$ and $K_{s p} o f A l (O H)_{3} = 1 \times 1 0^{- 33}$ ). Give answer by multiplying by $1 0^{- 16} .$

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A silver electrode is immersed in saturated $A g_{2} S O_{4 (a q)} .$ The potential difference between silver and the standard hydrogen electrode is found to be $0.711 V$ . Determine $K s p (A g_{2} S O_{4})$ .
[Given $E_{A g^{+} / A g}^{\circ} = 0.799 V]$

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A solution has $0.05 M$ $M g^{2 +}$ and $0.05 M$ $N H_{3}$ . The concentration of $N H_{4} C l$ required to prevent the formation of $M g (O H)_{2}$ in solution is:
$K_{s p}$ of $M g (O H)_{2} = 9.0 \times 10^{- 12}$ and $K_{b}$ of $N H_{3} = 1.8 \times 10^{- 5}$ .

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In a saturated solution of the sparingly soluble strong electrolyte $A g l O_{3}$ , (molecular mass = 283) the equilibrium which sets in is $A g l O_{3} (s) ⇌ A g^{+} (a q) + I O_{3}^{-} (a q)$ if the solubility product constant. $K_{s p}$ of $A g l O_{3}$ , at a given temperature is $1.0 \times 10^{- 8}$ , what is the mass of $A g l O_{3}$ , contained in lot) nil. of its saturated solution?

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At $- 5 0^{\circ} C$ , liquid $N H_{3}$ has ionic product is $1 0^{- 30}$ . How many amide $(N H_{2}^{-})$ ions are present per $m m^{3}$ is pure liquid $N H_{3}$ ? (Take $N_{A} = 6 \times 1 0^{23})$ .

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The solubility of $A g I$ in $N a I$ solution is less than that in pure water because:

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Alkyl halides are polar and immiscible with water. Explain.

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At $2 5^{o} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 1 0^{- 11}$ . At what $p H$ , will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of $0.001 M M g^{2 +}$ ions?

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Which of the following is most soluble?

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$0.00050$ mole of $N a H C O_{3}$ is added to a large volume of a solution buffered at $p H = 8.00$ . How much material will exist in each of the three forms, $H_{2} C O_{3}, H C O_{3}^{-}, C O_{3}^{2 -}$ . $K_{1}$ and $K_{2}$ for $H_{2} C O_{3}$ are $4.5 \times 10^{- 7}$ and $4.5 \times 10^{- 11}$ respectively.

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Solubility Equilibria

Overview

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Videos

Study Materials

Revise with Concepts

Solubility and Solubility Product of sparingly soluble salts

Common Ion Effect on Solubility

Related questions

What is the ratio of moles of $M g (O H)_{2}$ and $A l (O H)_{3}$ present in 1 lit saturated aqueous solution of $M g (O H)_{2}$ & $A l (O H)_{3} (K_{s p} o f M g (O H)_{2} = 4 \times 1 0^{- 12}$ and $K_{s p} o f A l (O H)_{3} = 1 \times 1 0^{- 33}$ ). Give answer by multiplying by $1 0^{- 16} .$

A silver electrode is immersed in saturated $A g_{2} S O_{4 (a q)} .$ The potential difference between silver and the standard hydrogen electrode is found to be $0.711 V$ . Determine $K s p (A g_{2} S O_{4})$ .
[Given $E_{A g^{+} / A g}^{\circ} = 0.799 V]$

A solution has $0.05 M$ $M g^{2 +}$ and $0.05 M$ $N H_{3}$ . The concentration of $N H_{4} C l$ required to prevent the formation of $M g (O H)_{2}$ in solution is:
$K_{s p}$ of $M g (O H)_{2} = 9.0 \times 10^{- 12}$ and $K_{b}$ of $N H_{3} = 1.8 \times 10^{- 5}$ .

At $- 5 0^{\circ} C$ , liquid $N H_{3}$ has ionic product is $1 0^{- 30}$ . How many amide $(N H_{2}^{-})$ ions are present per $m m^{3}$ is pure liquid $N H_{3}$ ? (Take $N_{A} = 6 \times 1 0^{23})$ .

The solubility of $A g I$ in $N a I$ solution is less than that in pure water because:

Alkyl halides are polar and immiscible with water. Explain.

At $2 5^{o} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 1 0^{- 11}$ . At what $p H$ , will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of $0.001 M M g^{2 +}$ ions?

Which of the following is most soluble?

More topics

Solubility Equilibria

Overview

Learn

Videos

Study Materials

Revise with Concepts

Solubility and Solubility Product of sparingly soluble salts

Common Ion Effect on Solubility

Related questions

What is the ratio of moles of Mg(OH)2​ and Al(OH)3​ present in 1 lit saturated aqueous solution of Mg(OH)2​ & Al(OH)3​(Ksp​ of Mg(OH)2​=4×10−12 and Ksp​ of Al(OH)3​=1×10−33). Give answer by multiplying by 10−16.

A silver electrode is immersed in saturated Ag2​SO4(aq)​. The potential difference between silver and the standard hydrogen electrode is found to be 0.711 V. Determine Ksp(Ag2​SO4​). [Given EAg+/Ag∘​=0.799 V]

A solution has 0.05M Mg2+ and 0.05M NH3​. The concentration of NH4​Cl required to prevent the formation of Mg(OH)2​ in solution is: Ksp​ of Mg(OH)2​=9.0×10−12 and Kb​ of NH3​=1.8×10−5.

At −50∘C, liquid NH3​ has ionic product is 10−30. How many amide (NH2−​) ions are present per mm3 is pure liquid NH3​? (Take NA​=6×1023).

The solubility of AgI in NaI solution is less than that in pure water because:

Alkyl halides are polar and immiscible with water. Explain.

At 25oC, the solubility product of Mg(OH)2​ is 1.0×10−11. At what pH, will Mg2+ ions start precipitating in the form of Mg(OH)2​ from a solution of 0.001 M Mg2+ ions?

Which of the following is most soluble?

0.00050 mole of NaHCO3​ is added to a large volume of a solution buffered at pH=8.00. How much material will exist in each of the three forms, H2​CO3​,HCO3−​,CO32−​. K1​ and K2​ for H2​CO3​ are 4.5×10−7 and 4.5×10−11 respectively.

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What is the ratio of moles of $M g (O H)_{2}$ and $A l (O H)_{3}$ present in 1 lit saturated aqueous solution of $M g (O H)_{2}$ & $A l (O H)_{3} (K_{s p} o f M g (O H)_{2} = 4 \times 1 0^{- 12}$ and $K_{s p} o f A l (O H)_{3} = 1 \times 1 0^{- 33}$ ). Give answer by multiplying by $1 0^{- 16} .$

A silver electrode is immersed in saturated $A g_{2} S O_{4 (a q)} .$ The potential difference between silver and the standard hydrogen electrode is found to be $0.711 V$ . Determine $K s p (A g_{2} S O_{4})$ .
[Given $E_{A g^{+} / A g}^{\circ} = 0.799 V]$

A solution has $0.05 M$ $M g^{2 +}$ and $0.05 M$ $N H_{3}$ . The concentration of $N H_{4} C l$ required to prevent the formation of $M g (O H)_{2}$ in solution is:
$K_{s p}$ of $M g (O H)_{2} = 9.0 \times 10^{- 12}$ and $K_{b}$ of $N H_{3} = 1.8 \times 10^{- 5}$ .

At $- 5 0^{\circ} C$ , liquid $N H_{3}$ has ionic product is $1 0^{- 30}$ . How many amide $(N H_{2}^{-})$ ions are present per $m m^{3}$ is pure liquid $N H_{3}$ ? (Take $N_{A} = 6 \times 1 0^{23})$ .

The solubility of $A g I$ in $N a I$ solution is less than that in pure water because:

At $2 5^{o} C$ , the solubility product of $M g (O H)_{2}$ is $1.0 \times 1 0^{- 11}$ . At what $p H$ , will $M g^{2 +}$ ions start precipitating in the form of $M g (O H)_{2}$ from a solution of $0.001 M M g^{2 +}$ ions?