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Li^(+) ion is the smallest one among th...

`Li^(+)` ion is the smallest one among the ions of group- 1 elements. It would, therefore, be expected to have much higher ionic mobility and hence the solutions of lithium salts would be expected to have higher conductivity than that of solutions of cesium salts. However in reality, the reverse is observed. Explain.

Text Solution

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Due to high charge density, very small `Li^(+)` ion gets much more hydrated compared to the large `Cs^(+)` ion. Thus, size of the hydrated lithium ion is much larger than that of the hydrated cesium ion. For this reason, the mobility of `Li^(+)` ion is much lower than that of `Cs^(+)` ion and consequently, the solutions of lithium salts has much lower conductivity than the solutions of cesium salts.
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Alkali metal salts are water soluble and ionic in nature. Two factors such as lattice energy and hydration energy decide the extent of solubility. These two factors are opposing in nature solubility will be high if hydration energy is high. Higher will be the extent of hydration if size of the ion is smaller. Hydration of ions is exothermic process.Hydration energy also plays a role in deciding reducing power of alkali metals in solution. Identify the strongest reducing agent among alkali metals in solutions

Knowledge Check

  • If a sparingly soluble salt is placed in water, after some time an equilibrium is established when the rate of dissolution of ions from the solid equals the rate of precipitation of ions from the saturated solution at a particular temperature. Thus, a dynamic equilibrium exists between the undissociated solid species and the dissolved ionic species ina saturated solution at a particular temperature. For example, in AgCl, we have the equilibrium AgCl iff Ag^+ (aq)+Cl(aq ) K_(eq) = ([Ag^+][Cl^-])/[AgCl] K_(eq)xx[AgCl]=[Ag^+][Cl^-]..........(A) K_(sp)(AgCl)=[Ag^+][Cl^-] Therefore [AgCl] is constant If there would not have been a saturated solution, then from equation (A), Keq. [AgCl] ne K_(sp) but K_(eq).[AgCl]=Q_(AgCl) where Q is ionic product. It implies that for a saturated solution,,br> Q=K_(sp) K_(sp) is temperature depended. When Q lt K_(sp) , then the solution is unsaturated and there will be no precipitate formation. When Q=K_(sp) , then solution will be saturated, no ppt, will be formed When Qgt K_(sp) , the solution will be supersaturated and there will be formation precipitate A solution is a mixture of 0.05 M NaCl and 0.05 M Nal. The concentration of iodide ion in the solution when AgCl just starts precipitating is equal to

    A
    `4xx10^-6 M`
    B
    `2xx10^-8 M`
    C
    `2xx10^-7 M`
    D
    `8xx10^-15 M`

  • If a sparingly soluble salt is placed in water, after some time an equilibrium is established when the rate of dissolution of ions from the solid equals the rate of precipitation of ions from the saturated solution at a particular temperature. Thus, a dynamic equilibrium exists between the undissociated solid species and the dissolved ionic species ina saturated solution at a particular temperature. For example, in AgCl, we have the equilibrium AgCl iff Ag^+(aq)+Cl(aq) K_(eq) = ([Ag^+][Cl^-])/[AgCl] K_(eq)xx[AgCl]=[Ag^+][Cl^-]..........(A) K_(sp)(AgCl)=[Ag^+][Cl^-] Therefore [AgCl] is constant If there would not have been a saturated solution, then from equation (A), Keq. [AgCl] ne K_(sp) but K_(eq).[AgCl]=Q_(AgCl) where Q is ionic product. It implies that for a saturated solution,,br> Q=K_(sp) K_(sp) is temperature depended. When Q lt K_(sp) , then the solution is unsaturated and there will be no precipitate formation. When Q=K_(sp) , then solution will be saturated, no ppt, will be formed When Q gt K_(sp) , the solution will be supersaturated and there will be formation precipitate Answer the following questions based on above passage: pH of a saturated solution of Ba(OH)_2 is 12. Hence K_(sp) of Ba(OH)_2 is

    A
    `5xx10^-7 M^3`
    B
    `5xx10^-4 M^2`
    C
    `1xx10^-6 M^3`
    D
    `4xx10^-6 M^3`

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    Alkali metal salts are water soluble and ionic in nature. Two factors such as lattice energy and hydration energy decide the extent of solubility. These two factors are opposing in nature solubility will be high if hydration energy is high. Higher will be the extent of hydration if size of the ion is smaller. Hydration of ions is exothermic process.Hydration energy also plays a role in deciding reducing power of alkali metals in solution. Among the following the least soluble is

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