A solution contains 25 mmol of H3PO4 and 10. mmol of NaH2PO4. What volume of 2.0 M NaOH must be added to reach the second equivalence point of the titration of the H3PO4 with NaOH?
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To find the volume of 2.0 M NaOH required to reach the second equivalence point of the titration, we first need to calculate the moles of acid present in the solution.
1. Calculate the moles of H3PO4:
Moles of H3PO4 = 25 mmol = 0.025 mol
2. Calculate the moles of NaH2PO4:
Moles of NaH2PO4 = 10 mmol = 0.010 mol
3. Determine the limiting reagent:
H3PO4 will react with NaOH in a 1:1 ratio, whereas NaH2PO4 will react with NaOH in a 2:1 ratio. Since NaH2PO4 has a lower stoichiometric coefficient, it will be exhausted first. Therefore, NaH2PO4 is the limiting reagent.
4. Calculate the moles of NaOH required to react with NaH2PO4:
Moles of NaOH = 2 * Moles of NaH2PO4 = 2 * 0.010 mol = 0.020 mol
5. Calculate the volume of 2.0 M NaOH needed:
Moles = Molarity * Volume (in liters)
Volume of NaOH = Moles of NaOH / Molarity of NaOH = 0.020 mol / 2.0 M = 0.01 L = 10 mL
Therefore, 10 mL of 2.0 M NaOH must be added to reach the second equivalence point of the titration.
To determine the volume of 2.0 M NaOH required to reach the second equivalence point, we need to consider the stoichiometry of the reaction between H3PO4 and NaOH.
First, let's write the balanced chemical equation for the reaction:
H3PO4 + NaOH → NaH2PO4 + H2O
According to the reaction, one mole of H3PO4 reacts with three moles of NaOH. This means that for every millimole (mmol) of H3PO4, we need to add three millimoles (mmol) of NaOH.
Given that the solution contains 25 mmol of H3PO4, we can calculate the amount of NaOH needed as follows:
Amount of NaOH (mmol) = 3 * Amount of H3PO4 (mmol)
= 3 * 25 mmol
= 75 mmol
Now let's convert the amount of NaOH into volume using the molarity (2.0 M) of the NaOH solution:
Volume of NaOH (L) = Amount of NaOH (mmol) / Molarity of NaOH (M)
= 75 mmol / 2.0 M
= 37.5 mL
Therefore, you would need to add 37.5 mL of 2.0 M NaOH to reach the second equivalence point of the titration.
Look at titrating the first H, then double it.
H3PO4 + NaOH ==> NaH2PO4 + H2O
mols H3PO4 = 25 mmols.
So you will need 25 mmols NaOH to react with the first H of H3PO4.
M NaOH = mols NaOH/L NaOH
2.0M = 25 mmols/mL
mL = 25/2.0 = 12.5 mL.
So it will take 25 to neutralize the second H of H3PO4 (and 37.5 mL to neutralize the 3rd H although you probably wouldn't know when you were there.)