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A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution....

A 0.2688 g sample of a monoprotic acid neutralized 16.4 mL of 0.08133 M KOH solution. Calculate the molar mass of the acid. (5 points)

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Answer #1

A monoprotic acid are the ones that produce 1 hydrogen H+ , it has a form of HX, x is unknown, it only needs 1 ion of OH-

HX + KOH H2O + KX

It is a 1:1 reaction, 1 mole of KOH will neutralize 1 mole of HX

the moles of KOH :

moles = Molarity * Volume

moles = 0.08133 M * 0.0164 L = 0.00133381 moles of KOH

So we also have 0.00133381 moles of acid

Since we know the grams of acid we apply the next formula

molar mass = 0.2688 grams / 0.00133381 moles = 201.527 g/gmol

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