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

Hint: 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.