How many milliliters of a 0.1 M NaOH solution is needed to completely neutralize 1 liter of lemonade containing 25% citric acide
25% is a lot of citric acid in lemonade, it is usually ca 5% or less.
Start from a balanced equation:
NaOH + 3-carboxy-3-hydroxypentanedioic acid -> 3-carboxy-3-hydroxypentanedioic acid, trisodium salt + H2O
which you need to balance
OK. Let us assume that the 25% is by weight and that the density 1.00 g L^-1
so our solution contains 250 g of citric acid, molar mass = 192 g mole^-1
so the number of moles =
250 g/192 g mole^-1
from the balanced equation we can work out how many moles of NaOH react with
1 mole of citric acid.
Hence work out how many moles (N) of NaOH react with 250 g/192 g mole^-1
of citric acid.
To calculate the volume (millitres) of NaOH required
=Nx1000/0.1
it will be a large volume.