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A solid sample of Zinc Hydroxide is added to 0.350 L of 0.500 M aqueous Hydrogen Bromide. The solution that remains is still acidic. It is then titrated with 0.500 M NaOH solution, and it takes 88.5 mL of the NaOH solution to reach the equivalence point. What mass of Zinc Hydroxide was added to the Hydrogen Bromide solution?

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Answer:

6.499 g

Explanation:

One part of the aqueous HBr reacted with Zinc Hydroxide following the reaction:

  • Zn(OH)₂ + 2HBr → ZnBr₂ + 2H₂O

And the remaining HBr reacted with NaOH:

  • NaOH + HBr → NaBr + H₂O

First we calculate how many HBr moles remained after reacting with Zn(OH)₂. That number equals the number of NaOH moles used in the titration:

  • 0.500 M * 88.5 mL = 44.25 mmol NaOH = mmol HBr

Now we calculate how many moles of HBr reacted with Zn(OH)₂:

  • Originally there were (350 mL * 0.500 M) 175 mmol HBr
  • 175 mmol - 44.25 mmol = 130.75 mmol HBr

Then we convert those 130.75 mmoles of HBr to the Zn(OH)₂ moles they reacted with:

  • 130.75 mmol HBr * [tex]\frac{1mmolZn(OH)_{2}}{2mmolHBr}[/tex] = 65.375 mmol Zn(OH)₂.

Finally we convert Zn(OH)₂ moles to grams:

  • 65.375 mmol Zn(OH)₂ * 99.424 mg/mmol = 6499.8 mg Zn(OH)₂
  • 6499.8 mg Zn(OH)₂ / 1000 = 6.499 g

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