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## Homework Statement

1.5000 g of diprotic weak acid H

_{2}A was dissolved in 100.00 mL volumetric flask.

25.00 mL aliqouts of this solution was titrated with a monoprotic strong base NaOH (0.08000 M). The titre volume of NaOH was 40.00 mL. Calculate the molecular weight of H

_{2}A.

## Homework Equations

N/A

## The Attempt at a Solution

1 mole of H

_{2}A contains 2 moles of H

^{+}

1 mole of NaOH contains 1 mole of OH

^{-}

2C

_{a}V

_{a}= C

_{b}V

_{b}

2 (C

_{a})(0.02500 L) = (0.0800 M)(0.04000L)

C

_{a}= 0.0640 M

#moles of H

_{2}A = C

_{a}V

_{a}

= (0.0640 M)(0.10000 L)

= 0.00640 mol

molecular mass of H

_{2}A = mass/# moles

= 1.5000 g / 0.00640 mol

= 234.375 g/mol

My answer is 234.375 but something seems odd about it. Can anyone tell me if I missed something or did something wrong? Thanks!

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