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Determine solubility of boric acid (H3BO3) at various pH's

This problem indicates that 1 gram of hydrated boric acid (H3BO3) is added to 1 liter of water at pH 7.0 and at standard temperature.

The solubility is stated in the problem as 5.7 grams per 100 ml of solution. A pKa is also given as 9.24.

1) What does the pKa value mean?
2) How would I determine the amount of boric acid that will dissolve at pH 7 given the conditions above?
3) And then how would I determine how much boric acid dissolved would dissolve if the pH were lowered to 6.0?

Solution Preview

1) pKa = -logKa
where Ka is an acid dissociation constant, which is an equilibrium constant for the dissociation of a weak acid. When an acid, HA, dissolves in water, some molecules of the acid "dissociate" to form hydronium ions and the conjugate base of the acid:
HA = H+ + A-
Thus Ka=[H+][A-]/[HA]

2) You're given solubility of 5.7 g in 100mL. This is the maximum amount of boric acid dissolving in water. Since you adding much less than this threshold (1g in 1L), the equilibrium will need to be ...

Solution Summary

The solution discusses properties of hydrated boric acid.