How many mL of 0.100 M NaOH are required to neutralize (or titrate) 25.0 mL of 0.15 M H3PO4?
Normalityacid*volumeacid=Normalitybase*Vb
vb=.15*3*25ml/.1=4.5*25ml
To determine the number of mL of 0.100 M NaOH required to neutralize 25.0 mL of 0.15 M H3PO4, we can use the concept of stoichiometry and the balanced equation for the reaction between NaOH and H3PO4.
The balanced equation for the reaction is as follows:
3NaOH + H3PO4 → Na3PO4 + 3H2O
From the balanced equation, we can see that 3 moles of NaOH react with 1 mole of H3PO4. This ratio allows us to establish a conversion factor.
Step 1: Convert the given volume of H3PO4 to moles.
Using the molarity (0.15 M) and volume (25.0 mL) of H3PO4, we can calculate the number of moles of H3PO4.
Moles of H3PO4 = Molarity × Volume (in L)
Moles of H3PO4 = 0.15 mol/L × 0.025 L
Moles of H3PO4 = 0.00375 mol
Step 2: Determine the number of moles of NaOH needed to neutralize the H3PO4.
Since the stoichiometric ratio of NaOH to H3PO4 is 3:1, we can establish the following conversion factor:
1 mole H3PO4 → 3 moles NaOH
Moles of NaOH = 3 × Moles of H3PO4
Moles of NaOH = 3 × 0.00375 mol
Moles of NaOH = 0.01125 mol
Step 3: Convert the number of moles of NaOH to volume (in mL).
Using the molarity (0.100 M) of NaOH, we can calculate the volume of NaOH required.
Volume (mL) of NaOH = Moles of NaOH ÷ Molarity (in M)
Volume (mL) of NaOH = 0.01125 mol ÷ 0.100 mol/L
Volume (mL) of NaOH = 0.1125 L × 1000 mL/L
Volume (mL) of NaOH = 11.25 mL
Therefore, 11.25 mL of 0.100 M NaOH are required to neutralize 25.0 mL of 0.15 M H3PO4.