Show your work and explain how you would find out how many liters of 0.500 M H3PO4 it would take to react with 8.00 L of 0.100 M NaOH.
Given molarity of H3PO4 is 0.5M, volume=? and NaOH Molarity=0.1 M and volume=8 L.
Now find the moles of NaOH=Molarity x volume=0.1 moles/Lx8 L=0.8 moles.
The balanced equation between H3PO4 and NaOH is
H3PO4+3NaOH------>Na3PO4+3H2O
Therefore mole ratio between H3PO4 and NaOH is 1:3.
So moles of H3PO4=1/3(moles of NaOH)=1/3(0.8 moles)=0.266 moles H3PO4.
Given Molarity of H3PO4=0.5 M, so volume=moles/Molarity
Volume=0.266 moles/0.5 moles/L=0.533 L
So volume of H3PO4 is 0.533 L.
(Or) We can find out in another way
First convert Molarity into normality.
So normality of H3PO4=Molarity X3(since it has 3H's)=0.5x3=1.5 N, and for NaOH Molarity=normality (since it has 1 OH )
Therefore for NaOH 0.1 N and 8 L volume
From dilution law, N1V1=N2V2
1.5 NxV1=0.1 Nx8 L
V1=0.8/1.5=0.533 L.
So volume of H3PO4 is 0.533 L.
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