Ksp of Borax

This experiment is a practical application of chemical equilibrium focused on determining the solubility product constant ($\text{K}_{\text{sp}}$) of a sparingly soluble ionic compound, borax ($\text{Na}_2\text{B}_4\text{O}_7 \cdot 10\text{H}_2\text{O}$). $\text{K}_{\text{sp}}$ is a key thermodynamic value that quantifies the extent to which a solid compound dissolves in water, thus defining its solubility. The determination relies on preparing a saturated solution and accurately measuring the concentration of one of its ions using the quantitative technique of titration.

Background

Solubility Equilibrium and $\text{K}_{\text{sp}}$

When a sparingly soluble ionic compound like borax is placed in water, it establishes a heterogeneous equilibrium between the undissolved solid and its dissolved ions. The dissolution reaction for borax in water is often simplified to its anhydrous form for the equilibrium expression:

$$\text{Na}_2\text{B}_4\text{O}_7(\text{s}) \rightleftharpoons 2\text{Na}^+(\text{aq}) + \text{B}_4\text{O}_7^{2-}(\text{aq})$$

The $\text{K}_{\text{sp}}$ expression for this equilibrium is defined by the product of the concentrations of the dissolved ions, each raised to the power of its stoichiometric coefficient:

$$\text{K}_{\text{sp}} = [\text{Na}^+]^2[\text{B}_4\text{O}_7^{2-}]$$

Because borax is the only source of the ions, we can define the molar solubility of borax as $s = [\text{B}_4\text{O}_7^{2-}]$. Since two $\text{Na}^+$ ions are produced for every one $\text{B}_4\text{O}_7^{2-}$ ion, the equilibrium concentrations are:

$$[\text{B}_4\text{O}_7^{2-}] = s \quad \text{and} \quad [\text{Na}^+] = 2s$$

Substituting these into the $\text{K}_{\text{sp}}$ expression yields:

$$\text{K}_{\text{sp}} = (2s)^2(s) = 4s^3$$

The primary goal of the lab is to find the value of $s = [\text{B}_4\text{O}_7^{2-}]$ through titration.

Titration of Tetraborate Ion ($\text{B}_4\text{O}_7^{2-}$)

The concentration of the tetraborate ion ($\text{B}_4\text{O}_7^{2-}$) in the saturated solution is determined by an acid-base titration. The tetraborate ion acts as a weak base, which is titrated using a standardized strong acid ($\text{HCl}$).

The reaction for the titration is:

$$\text{B}_4\text{O}_7^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 5\text{H}_2\text{O}(\text{l}) \rightarrow 4\text{H}_3\text{BO}_3(\text{aq})$$

By accurately measuring the volume of standardized $\text{HCl}$ required to reach the equivalence point, the moles of $\text{H}^+$ are known. Using the stoichiometry (2 moles of $\text{H}^+$ react with 1 mole of $\text{B}_4\text{O}_7^{2-}$), the concentration of $\text{B}_4\text{O}_7^{2-}$ (which equals $s$) can be calculated. This value, along with the initial temperature of the solution (which affects $s$ and must be measured), allows for the final calculation of $\text{K}_{\text{sp}}$.