What volume of 0.350 m koh is required to react completely with 24.0 ml of 0.650 m h3po4?

The complete balanced chemical equation for this is:

3KOH  +  H3PO4  -->  K3PO4  +  3H2O

First we calculate the number of moles of H3PO4:

moles H3PO4 = 0.650 moles / L * 0.024 L = 0.0156 mol

From stoichiometry, 3 moles of KOH is required for every mole of H3PO4, therefore:

moles KOH = 0.0156 mol H3PO4 * (3 moles KOH / 1 mole H3PO4) = 0.0468 mol

Calculating for volume given molarity of 0.350 M KOH:

Volume = 0.0468 mol / (0.350 mol / L) = 0.1337 L = 133.7 mL

133.7 mL KOH

nacl is added to the aqueous of what? ? in order to get a proper answer you have to specify the exact composition of the aqueous solution.

1. they vary with a higher number because boiling is way hotter than melting