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

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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?

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Solution Summary

The solution discusses properties of hydrated boric acid.

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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 ...

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