A titration involves titrating 0.4 M NaOH into a solution of 0.2 M H3PO4.
a.) Calculate the volume of NaOH that will be required to reach the first equivalence point.
b.) Calculate the volume of NaOH required to reach the second equivalence point.
Other information: 10 ml H3PO4 was added to 250 ml of deionized water, and solution was then titrated with NaOH.
Is that 10 mL of the 0.2M H3PO4 added to 250 mL H2O to make 260 mL solution? If you titrate the entire solution, which seems unlikely, then you will have
H3PO4 ==> NaH2PO4 + H2O
mols H3PO4 = 0.010L x 0.2M = 0.002
mols NaOH required = 0.002 since 1 mol H3PO4 = 1 mol NaOH.
Then M NaOH = mols NaOH/L NaOH or
L NaOH = mols NaOH/M NaOH = 0.002/0.4 = approx 0.005L or 5 mL
For part B you are titrating the second H do it will take twice that or abouat 10 mL.
If I've interpreter the problem incorrectly please rephrase the question so it is clear what is being titrated.
Oh, sorry. It was supposed to be 50 ml of water instead of 250 ml.
To determine the volume of NaOH required to reach the first equivalence point, you need to know the stoichiometry of the reaction between NaOH and H3PO4.
The balanced equation for the reaction is:
3 NaOH + H3PO4 -> Na3PO4 + 3 H2O
From the balanced equation, you can see that the stoichiometric ratio between NaOH and H3PO4 is 3:1. This means that for every 3 moles of NaOH, you will react with 1 mole of H3PO4.
To calculate the volume of NaOH needed, you can use the following equation:
(Moles of H3PO4) = (Volume of NaOH) x (Molarity of NaOH) = (Moles of NaOH) x (Molarity of H3PO4)
Since the molarity and volume of H3PO4 are known, you can rearrange the equation to solve for the volume of NaOH:
Volume of NaOH = (Moles of H3PO4) / (Molarity of NaOH)
a.) To calculate the volume of NaOH required to reach the first equivalence point, you need to know the number of moles of H3PO4 at the equivalence point. The first equivalence point occurs when all the moles of H3PO4 have reacted with the moles of NaOH in a 1:3 ratio.
Moles of H3PO4 = (Molarity of H3PO4) x (Volume of H3PO4)
Since 10 mL of 0.2 M H3PO4 was added to 250 mL of water, the total volume of H3PO4 solution is 260 mL (10 mL + 250 mL). You can convert this to liters:
Volume of H3PO4 = 260 mL / 1000 = 0.26 L
Now you can calculate the moles of H3PO4:
Moles of H3PO4 = (0.2 M) x (0.26 L) = 0.052 moles
To reach the first equivalence point, 1 mole of H3PO4 reacts with 3 moles of NaOH. Therefore, the moles of NaOH required are:
Moles of NaOH = 3 x (Moles of H3PO4) = 3 x 0.052 = 0.156 moles
Finally, you can calculate the volume of NaOH required:
Volume of NaOH = (0.156 moles) / (0.4 M) = 0.39 L
b.) To calculate the volume of NaOH required to reach the second equivalence point, you need to know the number of moles of H3PO4 at the second equivalence point. The second equivalence point occurs when all the moles of H3PO4 have reacted with the moles of NaOH in a 1:3 ratio, but this time there is twice as much H3PO4 compared to the first equivalence point.
Moles of H3PO4 = 2 x (0.052 moles) = 0.104 moles
Using the same equation as before, you can calculate the volume of NaOH required:
Volume of NaOH = (0.104 moles) / (0.4 M) = 0.26 L
So, the volume of NaOH required to reach the first equivalence point is 0.39 L, and the volume of NaOH required to reach the second equivalence point is 0.26 L.