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A : Precipitates formation takes place w...

A : Precipitates formation takes place when
`K_(ip) gt K_(sp)` .
R : `K_(ip) = K_(sp)` for a saturated solution.

A

If both Assertion & Reason are true and the reason in the corret explanation of the assertion, then mark

B

If both assertion & Reason are true but the reason is not the correct explanation of the assertion, then mark

C

If Assertion is true statement but Reason is false then mark

D

If both Assertion and Reason are false statements , then mark

Text Solution

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The correct Answer is:
To solve the question, we need to analyze the assertion and the reason provided, and determine their truth values and relationships. ### Step 1: Analyze the Assertion The assertion states: "Precipitates formation takes place when \( K_{ip} > K_{sp} \)." - **Explanation**: The ionic product (\( K_{ip} \)) is the product of the concentrations of the ions in solution. The solubility product constant (\( K_{sp} \)) is a specific type of equilibrium constant that applies to the solubility of ionic compounds. When \( K_{ip} > K_{sp} \), it indicates that the solution is supersaturated, leading to the formation of precipitates as the system attempts to reach equilibrium. - **Truth Value**: This assertion is **True**. ### Step 2: Analyze the Reason The reason states: "\( K_{ip} = K_{sp} \) for a saturated solution." - **Explanation**: When a solution is saturated, the concentrations of the ions in the solution are at equilibrium with the solid phase. At this point, the ionic product equals the solubility product constant, indicating that no further dissolution or precipitation occurs. - **Truth Value**: This reason is also **True**. ### Step 3: Determine the Relationship Now, we need to determine whether the reason is the correct explanation for the assertion. - The assertion explains a condition for precipitate formation, while the reason describes the condition for a saturated solution. Although both statements are true, the reason does not explain why precipitates form when \( K_{ip} > K_{sp} \); it merely states a condition of equilibrium. ### Conclusion Since both the assertion and reason are true, but the reason does not correctly explain the assertion, the correct option is: **Answer: Option B** - Both assertion and reason are true, but the reason is not the correct explanation of the assertion. ---
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The product of the concentration of the ions of an electrolyte raised to power of their coefficients in the balanced chemical equation in the solution at any concentration . Its value is not constant and varies with change in concentration . Its value is not constant and varies with change in concentration . Its value is not constant and varies with change in concentration . Ionic product of the saturated solution is called solubility product K_(sp) (i) When K_(ip) = K_(sp) , the solution is just saturated and no precipitation takes place . (ii) When K_(ip) lt K_(sp) the solution is unsaturated and precipitation will not take place . (iii) When K_(ip) gt K_(sP) the solution is supersaturated and precipitation takes place . The solubility product K_(sp) , of sparingly soluable salt MX at 25^(@) is 2.5 xx10^(-9) . The solubility of the salt in mol L^(-1) at this temperature is :

The product of the concentration of the ions of an electrolyte raised to power of their coefficients in the balanced chemical equation in the solution at any concentration . Its value is not constant and varies with change in concentration . Its value is not constant and varies with change in concentration . Its value is not constant and varies with change in concentration . Ionic product of the saturated solution is called solubility product K_(sp) (i) When K_(ip) = K_(sp) , the solution is just saturated and no precipitation takes place . (ii) When K_(ip) lt K_(sp) the solution is unsaturated and precipitation will not take place . When K_(ip) gt K_(sP) the solution is supersaturated and precipitation takes place . The concentration of Ag^(+) ions in a given saturated solution of AgCl at 25^(@)C is 1.06 xx10^(-5) g ion per litre . The solubility product of AgCl is :

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Determine the molar solubility of MgF_(2) from its solubility product K_(sp)=4xx10^(-9) :

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from At 25^(@)C, will a precipitate of Mg (OH)_(2) from when a 0.0001 M solution of Mg(NO_(3))_(2) is adjusted to a pH of 9.0 ? At what minimum value of pH will precipition start ? ["Given" : K_(sp)(Mg(OH)_(2))=10^(-11)M^(3)]

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Will a precipitate from if 1 volume of 0.1 volume of 0.1 MPb^(2+) ion solution in mixed with 3 volume of 0.3 M Cl^(-) ion solution ? ["Givem":K_(sp)(PbCl_(2))=1.7xx10^(-5)M^(3)]

Consider a sturated solution of silver chloride that is in contact with solid silver chloride. The solubility equilibrium can be represented as AgCl(s)hArrAg^(+)(aq.)+Cl^(-)(aq.)," "K_(sp)=[Ag^(+)(aq.)][Cl^(-)(aq.)] Where K_(sp) is clled the solubility product constant or simply the solubility product. In general, the solubility product of a compound is the product of the molar concentrations of the constituent ions, each raised to the power of its stoichiometric coefficient in the equilibrium equation. For concentrations of ions that do not necessarliy correpond to equilibrium conditions we use the reaction quotient (Q) which is clled the ion or ionic prodect (Q) to predict whether a precipitate will from. Note that (Q) has the same for as K_(sp) are QltK_(sp) Unsaturated solution Q=K_(sp) Saturated solution Qgt_(sp) Supersaturated solution, precipitate will from Will a precipitate from if 50 cm^(3) of 0.01 M AgNO_(3) and 50 cm^(3) of 2xx10^(-5) M NaCl are mixed? ["Given": K_(sp)(AgCl)=10^(-10)M^(2)]

A solution constains Zn^(2+) ions and Cu^(2+) ions each of 0.02M . If the solution is made 1M in H^(o+) , and H_(2)S is passed untill the solution is satured, should a precipitate be formed? Given: K_(sp) ZnS = 10^(-22) , K_(sp) Cus = 8 xx 10^(-37) . In satured solution, K_(sp) (H_(2)S) = 10^(-22)

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