Question

How many kilograms of salt (NaCl) can be dissolved in 2 liters of water at 25 degrees Celsius? The solubility product constant, ${{{K}}_{{{sp}}}}$ , is ${{38}}{{.65}}$ .

Answer 1

In the above question, we have to find out the amount of salt dissolved in 2 liters of water at 25 degrees celsius. We have to write a balanced chemical equation for the dissociation of NaCl salt. From ${{{K}}_{{{sp}}}}$ value, we can find the concentration of both the ions. Then we have to calculate the molar mass and convert the concentration into mass of NaCl.

Formula Used
For an equation:
${{AB}} \rightleftharpoons {{{A}}^{{ + }}}{{ + }}{{{B}}^{{ - }}}$
The value of ${{{K}}_{{{sp}}}}{{ = }}\left[ {{{{A}}^{{ + }}}} \right]\left[ {{{{B}}^{{ - }}}} \right]$

Complete step by step solution
In the above question, ${{{K}}_{{{sp}}}}$ value is given. So, let us first write how NaCl dissociates in water.
${{NaCl}} \rightleftharpoons {{N}}{{{a}}^{{ + }}}{{ + C}}{{{l}}^{{ - }}}$ ............................(1)
Since, when NaCl is dissociated, it forms 1 mole of sodium and chloride ions, which implies that concentration of both the ions is the same.
Let $\left[ {{{N}}{{{a}}^{{ + }}}} \right]{{ = }}\left[ {{{C}}{{{l}}^{{ - }}}} \right]{{ = x}}$
We know that ${{{K}}_{{{sp}}}}{{ = }}\left[ {{{N}}{{{a}}^{{ + }}}} \right]\left[ {{{C}}{{{l}}^{{ - }}}} \right]$
So, ${{{x}}^{{2}}}{{ = 38}}{{.65}}$
Which implies that ${{x = }}\sqrt {{{38}}{{.65}}} {{ = 6}}{{.21}}$
Therefore, $\left[ {{{N}}{{{a}}^{{ + }}}} \right]{{ = }}\left[ {{{C}}{{{l}}^{{ - }}}} \right]{{ = 6}}{{.21}}$ mol/L
From equation 1, we can say infer that 1 mole of NaCl gives 1 mole of sodium and chloride ions.
Since, the concentration of sodium and chloride ion is ${{6}}{{.21}}$ mol/L. Hence, concentration of NaCl is ${{6}}{{.21}}$ mol/L.
Molar mass of NaCl = atomic mass of Na + atomic of Cl = ${{23 + 35}}{{.5 = 58}}{{.5}}$ g
So, in 1 mole of NaCl, weight = ${{58}}{{.5}}$ g
So, in ${{6}}{{.21}}$ mol/L, weight of NaCl is ${{58}}{{.5 \times 6}}{{.21 = 363}}{{.2}}$ g/L
Amount of NaCl in 2 liters = ${{363}}{{.2 \times 2 = 726}}{{.4}}$ g = ${{0}}{{.72}}$ kg.
Therefore, ${{0}}{{.72}}$ kg of NaCl can be dissolved in 2 liters of water.

Note
The solubility product constant, ${{{K}}_{{{sp}}}}$ ​can be defined as the equilibrium constant for a solid substance which gets dissolves in an aqueous solution. It represents the level at which a solute dissolves in solution. The more soluble a substance is, the higher the ${{{K}}_{{{sp}}}}$ value it has.