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

Oct 28, 2015

The simple answer is with ease. But do you know what $\text{saturated}$ means?

#### Explanation:

At room temperature, about 10 g of $N a H C {O}_{3}$ 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:

$N a H C {O}_{3} \left(s\right) r i g h t \le f t h a r p \infty n s N a H C {O}_{3} \left(a q\right)$. As with any equilibrium we write the the equilibrium equation:

${K}_{s p} = \frac{\left[N a H C {O}_{3} \left(a q\right)\right]}{\left[N a H C {O}_{3} \left(s\right)\right]}$.

As you know, the term $\left[N a H C {O}_{3} \left(s\right)\right]$ is meaningless; you cannot speak of the concentration of a solid; so it is removed from the equation:

${K}_{s p} = \left[N a H C {O}_{3} \left(a q\right)\right] = \left[N {a}^{+} \left(a q\right)\right] \left[H C {O}_{3} \left(a q\right)\right]$. 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 $\left[N {a}^{+} \left(a q\right)\right] \left[H C {O}_{3} \left(a q\right)\right]$ would be greater than ${K}_{s p}$ and sodium bicarbonate would precipitate from solution.

${K}_{s p}$, the solubility product, is tabulated for a host of sparingly soluble and insoluble salts. They must be measured.