How many mL of .200 M NaOH are required to completely neutralize 5.00 mL of 0.100 M H2PO4?
To find the number of mL of NaOH required to neutralize 5.00 mL of H2PO4, we need to use the concept of stoichiometry and the balanced chemical equation between NaOH and H2PO4.
The balanced equation is as follows:
2 NaOH + H2PO4 -> Na2HPO4 + 2 H2O
From the equation, we can see that each mole of H2PO4 requires 2 moles of NaOH to react completely.
First, we need to calculate the number of moles of H2PO4 in the given volume:
moles of H2PO4 = volume in liters * concentration
= 5.00 mL * (1 L / 1000 mL) * 0.100 M
= 0.0005 moles
Since the stoichiometric ratio between NaOH and H2PO4 is 2:1, we need half that number of moles of NaOH:
moles of NaOH = (1/2) * moles of H2PO4
= (1/2) * 0.0005 moles
= 0.00025 moles
Now, we can calculate the volume of NaOH required using the equation:
volume of NaOH = moles of NaOH / concentration
= 0.00025 moles / 0.200 M
= 0.00125 L
Finally, convert the volume from liters to milliliters:
volume of NaOH = 0.00125 L * (1000 mL / 1 L)
= 1.25 mL
Therefore, 1.25 mL of 0.200 M NaOH is required to completely neutralize 5.00 mL of 0.100 M H2PO4.
See your other problem below but note you have made a typo on H2PO4.
You mean H3PO4 or NaH2PO4 or H2PO4^-.