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A solution containing 1.22 g of a diprotic acid H2CH2O4 (malonic acid)
was titrated with 45.5 mL of NaOH to reach the second equivalence
point. What is the concentration of the NaOH solution? (MW_malonic
acid = 104.06 g/mol)

Respuesta :

maacastrobr

Answer:

0.515 M

Explanation:

The reaction that takes place is:

  • H₂CH₂O₄ + 2NaOH → Na₂CH₂O₄ + 2H₂O

First we convert 1.22 g of malonic acid into moles, using its molar mass:

  • 1.22 g ÷ 104.06 g/mol = 0.01172 mol

Then we convert 0.01172 malonic acid moles into NaOH moles, using the stoichiometric coefficients of the balanced reaction:

  • 0.01172 mol H₂CH₂O₄ * [tex]\frac{2molNaOH}{1molH_2CH_2O_4}[/tex] = 0.02344 mol NaOH

Finally we calculate the concentration of the NaOH solution, using the number of moles and given volume:

Converting 45.5 mL ⇒ 45.5/1000 = 0.0455 L

  • 0.02344 mol / 0.0455 L = 0.515 M

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