[Bakergirl350]

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?

[yellowjacket]

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
hey