
“ While comparison of $L{i^ + }$and $N{a^ + }$ salts in terms of solubility of $CO_3^{2 - }, PO_4^{3 - }$ and ${F^ - }$ in ${H_2}O$ we can say that $L{i_2}C{O_3}, L{i_3}P{O_4}, LiF$ are soluble while sodium salts are insoluble."
Answer whether the above statement is true or false. If true enter 1, else enter 0.
Answer
571.2k+ views
Hint: The solubility of Ionic compound depends on the lattice energy and hydration energy. If the hydration energy is more than lattice energy then ionic compounds are soluble in water.
Complete answer:
> The energy required for the splitting of the ionic compound into ions is called lattice energy and energy is given out when the ions get hydrated is called hydration energy. In the given question Li and Na are alkali - metal and they form ionic bonds.
> Lattice energy depends on the size of the cation. Larger the size of cation small the lattice energy means down the group lattice energy decrease hence the solubility increase so salt of $N{a^ + }$ like $N{a_2}C{O_3}, N{a_3}P{O_4} , NaF$ dissolve in ${H_2}O$. But due to the high lattice energy of $L{i^ + }$. The compounds of Li like $L{i_2}C{O_3}$ and $L{i_3}P{O_4}$ cannot dissolve. Hence the statement is false.
So the correct answer is 0 because the above statement is false.
Note: Solubility is a property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.
Complete answer:
> The energy required for the splitting of the ionic compound into ions is called lattice energy and energy is given out when the ions get hydrated is called hydration energy. In the given question Li and Na are alkali - metal and they form ionic bonds.
> Lattice energy depends on the size of the cation. Larger the size of cation small the lattice energy means down the group lattice energy decrease hence the solubility increase so salt of $N{a^ + }$ like $N{a_2}C{O_3}, N{a_3}P{O_4} , NaF$ dissolve in ${H_2}O$. But due to the high lattice energy of $L{i^ + }$. The compounds of Li like $L{i_2}C{O_3}$ and $L{i_3}P{O_4}$ cannot dissolve. Hence the statement is false.
So the correct answer is 0 because the above statement is false.
Note: Solubility is a property referring to the ability for a given substance, the solute, to dissolve in a solvent. It is measured in terms of the maximum amount of solute dissolved in a solvent at equilibrium. The resulting solution is called a saturated solution.
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