How do we make a #"saturated solution"# of #"sodium bicarbonate"# with a #100*mL# volume of water?

1 Answer
Oct 28, 2015

The simple answer is with ease. But do you know what #"saturated"# means?

Explanation:

At room temperature, about 10 g of #NaHCO_3# will dissolve in 100 mL of water. (Cleary, this solubility will alter should the temperature change, which is why I must specify a temperature.)

Now saturation is an equilibrium property, which we may represent by the following equation:

#NaHCO_3(s) rightleftharpoons NaHCO_3(aq)#. As with any equilibrium we write the the equilibrium equation:

#K_(sp) = [[NaHCO_3(aq)]]/[[NaHCO_3(s)]]#.

As you know, the term #[NaHCO_3(s)]# is meaningless; you cannot speak of the concentration of a solid; so it is removed from the equation:

#K_(sp) = [NaHCO_3(aq)] = [Na^+(aq)][HCO_3(aq)]#. So if solid sodium carbonate is present, this equilibrium operates, sodium carbonate is saturated. If we added extra sodium salt (from whatever source), the ion product #[Na^+(aq)][HCO_3(aq)]# would be greater than #K_(sp)# and sodium bicarbonate would precipitate from solution.

#K_(sp)#, the solubility product, is tabulated for a host of sparingly soluble and insoluble salts. They must be measured.