[Nanothree]

A 7.500-g sample of a monoprotic weak acid, HA, is added to enough distilled water to produce 500.0 mL of a solution with pH = 2.716. The solution is then titrated with NaOH solution. At the point where the acid is half titrated, i.e., [HA] = [A-], the pH = 4.602. What is the molar mass of this acid?

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I recognize that at the midpoint, pH=pKa, so the pKa for this acid is 4.602. Now what?

[sdekivit]

now the following equilibrium will be present:

HA + H₂O <--> A^- + H₃O⁺

with Ka = [A^-][H₃O⁺]/[HA] = 10^-4.602 = 2.5 x 10^-5

now you know the pH of the begin solution (500 mL 7.5 g acid dissolved)

--> when the acid dissociates, x mol A^- and x mol H₃O⁺ will be formed and the same amount has reacted yielding the following equation:

Ka = x² / ([HA]) = 2.5 x 10^-5

and solve fow [HA] (assume the dissociation of the weak acid is neglectable) and calculate the molar mass.

[AWK]

Note, from titration you can find pKa!