Question

Use the reaction: 3 NaOH + H3PO4 = Na3PO4 + 3 H2O
B) How many mL of 0.385 M H3PO4 react with 50.0 mL of 0.404 M NaOH?
C) What is the maximum amount of Na3PO4 that is formed from the reaction of 25.00mL of 0.1050 M NaOH and 15.00 mL of 0.08650 M H3PO4?
D) What is the molarity of Na3PO4 in the solution prepared in c?

Answer 1

You can use this formula:

\[M_{acid}V_{acid} = M_{base}V_{base}\]


where m = molarity of the acid Moles/Liter
and V = the volume of acid used in Liters L

so we're asked to find the number of liters of H3PO4 that react with a given number of moles and volume of base.


so we re-arrange the formula for the volume of the acid which is what we're looking for.

\[\frac{ M_{base}V_{base} }{ M_{acid} } = V_{acid}\]

\[\frac{ (0.40\frac{ \cancel\Mol }{ \cancel\L })*(0.05L) }{ 0.385 (\frac{ \cancel\Mol }{ \cancel\L }) } = 0.052~L acid \]

Answer 2

@cuanchi do we need to take into account the number of moles in the molar ratio since it's not a 1:1 reaction