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Acid/ Base
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Topic: Acid/ Base (Read 175 times)
Bakergirl350
Labrat
Posts: 1
Acid/ Base
«
on:
November 02, 2009, 04:20:29 PM »
A sample of Carboxylic acid is added to 50mL of 2.27M NaOH. The excess NaOH required 28.7mL of 1.86M HCl for neutralization. How many mols of NaOH reacted with carboxylic acid?
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yellowjacket
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Posts: 8
Re: Acid/ Base
«
Reply #1 on:
November 02, 2009, 08:05:22 PM »
Start with the number of moles of NaOH originally present:
2.27M NaOH *0.050L = 0.114mol NaOH
Then calculate the excess moles of NaOH:
1.86M HCl * 0.0287L * (1mol NaOH/ 1mol HCl) = 0.0534mol NaOH
Now, subtract the moles left over after the reaction from the total number of moles to calculate the number of moles that reacted with carboxylic acid:
0.114mol original - 0.053mol excess = 0.061mol NaOH that reacted with carboxylic acid
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